添加两个二进制字符串的程序-yiteyi-C++库

给定两个二进制字符串，返回它们的和（也是一个二进制字符串）。 例子：

null

Input:  a = "11", b = "1"Output: "100"

我们强烈建议您尽量减少浏览器，并先自己尝试 其思想是从两个字符串的最后一个字符开始，逐个计算数字和。如果总和大于1，则存储下一位的进位。

C++

                         // C++ program to add two binary strings                       
                         #include <bits/stdc++.h>                       
                         using                                     namespace                                     std;                       
           
                         // This function adds two binary strings and return                       
                         // result as a third string                       
                         string addBinary(string A, string B)                       
                         {                       
                                                 // If the length of string A is greater than the length                       
                                                 // of B then just swap the the string by calling the                       
                                                 // same function and make sure to return the function                       
                                                 // otherwise recursion will occur which leads to                       
                                                 // calling the same function twice                       
                                                 if                                     (A.length() > B.length())                       
                                                 return                                     addBinary(B, A);                       
           
                                                 // Calculating the differnce between the length of the                       
                                                 // two strings.                       
                                                 int                                     diff = B.length() - A.length();                       
           
                                                 // Initialise the padding string which is used to store                       
                                                 // zeroes that should be added as prefix to the string                       
                                                 // which has length smaller than the other string.                       
                                                 string padding;                       
                                                 for                                     (                                     int                                     i = 0; i < diff; i++)                       
                                                 padding.push_back(                                     '0'                                     );                       
           
                                                 A = padding + A;                       
                                                 string res;                       
                                                 char                                     carry =                                     '0'                                     ;                       
           
                                                 for                                     (                                     int                                     i = A.length() - 1; i >= 0; i--) {                       
                                                 // This if condition solves 110 111 possible cases                       
                                                 if                                     (A[i] ==                                     '1'                                     && B[i] ==                                     '1'                                     ) {                       
                                                 if                                     (carry ==                                     '1'                                     )                       
                                                 res.push_back(                                     '1'                                     ), carry =                                     '1'                                     ;                       
                                                 else                       
                                                 res.push_back(                                     '0'                                     ), carry =                                     '1'                                     ;                       
                                                 }                       
                                                 // This if condition solves 000 001 possible cases                       
                                                 else                                     if                                     (A[i] ==                                     '0'                                     && B[i] ==                                     '0'                                     ) {                       
                                                 if                                     (carry ==                                     '1'                                     )                       
                                                 res.push_back(                                     '1'                                     ), carry =                                     '0'                                     ;                       
                                                 else                       
                                                 res.push_back(                                     '0'                                     ), carry =                                     '0'                                     ;                       
                                                 }                       
                                                 // This if condition solves 100 101 010 011 possible                       
                                                 // cases                       
                                                 else                                     if                                     (A[i] != B[i]) {                       
                                                 if                                     (carry ==                                     '1'                                     )                       
                                                 res.push_back(                                     '0'                                     ), carry =                                     '1'                                     ;                       
                                                 else                       
                                                 res.push_back(                                     '1'                                     ), carry =                                     '0'                                     ;                       
                                                 }                       
                                                 }                       
           
                                                 // If at the end their is carry then just add it to the                       
                                                 // result                       
                                                 if                                     (carry ==                                     '1'                                     )                       
                                                 res.push_back(carry);                       
                                                 // reverse the result                       
                                                 reverse(res.begin(), res.end());                       
           
                                                 // To remove leading zeroes                       
                                                 int                                     index = 0;                       
                                                 while                                     (index + 1 < res.length() && res[index] ==                                     '0'                                     )                       
                                                 index++;                       
                                                 return                                     (res.substr(index));                       
                         }                       
           
                         // Driver program                       
                         int                                     main()                       
                         {                       
                                                 string a =                                     "1101"                                     , b =                                     "100"                                     ;                       
                                                 cout << addBinary(a, b) << endl;                       
                                                 return                                     0;                       
                         }                       

JAVA

                         // java program to add                       
                         // two binary strings                       
           
                         public                                     class                                     GFG {                       
           
                                                 // This function adds two                       
                                                 // binary strings and return                       
                                                 // result as a third string                       
                                                 static                                     String addBinary(String a, String b)                       
                                                 {                       
                                   
                                                 // Initialize result                       
                                                 StringBuilder result =                                     new                                     StringBuilder(                                     ""                                     );                       
                                   
                                                 // Initialize digit sum                       
                                                 int                                     s =                                     0                                     ;                       
           
                                                 // Traverse both strings starting                       
                                                 // from last characters                       
                                                 int                                     i = a.length() -                                     1                                     , j = b.length() -                                     1                                     ;                       
                                                 while                                     (i >=                                     0                                     || j >=                                     0                                     || s ==                                     1                                     )                       
                                                 {                       
                                   
                                                 // Comput sum of last                       
                                                 // digits and carry                       
                                                 s += ((i >=                                     0                                     )? a.charAt(i) -                                     '0'                                     :                                     0                                     );                       
                                                 s += ((j >=                                     0                                     )? b.charAt(j) -                                     '0'                                     :                                     0                                     );                       
           
                                                 // If current digit sum is                       
                                                 // 1 or 3, add 1 to result                       
                                                 result.append((                                     char                                     )(s %                                     2                                     +                                     '0'                                     ));                       
           
                                                 // Compute carry                       
                                                 s /=                                     2                                     ;                       
           
                                                 // Move to next digits                       
                                                 i--; j--;                       
                                                 }                       
                                   
                                                 // Remove leading zeros, if any                       
                                                 int                                     start = result.length()-                                     1                                     ;                       
                                   
                                                 while                                     (start >=                                     0                                     && result.charAt(start) ==                                     '0'                                     ) {                       
                                                 start--;                       
                                                 }                       
                                   
                                                 if                                     (start != result.length()-                                     1                                     ) {                       
                                                 result.delete(start+                                     1                                     ,result.length());                       
                                                 }                       
                                   
                                                 return                                     result.reverse().toString();                       
                                                 }                       
           
                                                 //Driver code                       
                                                 public                                     static                                     void                                     main(String args[])                       
                                                 {                       
                                                 String a =                                     "1101"                                     , b=                                     "100"                                     ;                       
                                   
                                                 System.out.print(addBinary(a, b));                       
                                                 }                       
                         }                       
           
                         // This code is contributed by Sam007.                       

Python3

                         # Python Solution for above problem:                       
           
                         # This function adds two binary                       
                         # strings return the resulting string                       
                         def                                     add_binary_nums(x, y):                       
                                                 max_len                                     =                                     max                                     (                                     len                                     (x),                                     len                                     (y))                       
           
                                                 x                                     =                                     x.zfill(max_len)                       
                                                 y                                     =                                     y.zfill(max_len)                       
                                   
                                                 # initialize the result                       
                                                 result                                     =                                     ''                       
                                   
                                                 # initialize the carry                       
                                                 carry                                     =                                     0                       
           
                                                 # Traverse the string                       
                                                 for                                     i                                     in                                     range                                     (max_len                                     -                                     1                                     ,                                     -                                     1                                     ,                                     -                                     1                                     ):                       
                                                 r                                     =                                     carry                       
                                                 r                                     +                                     =                                     1                                     if                                     x[i]                                     =                                     =                                     '1'                                     else                                     0                       
                                                 r                                     +                                     =                                     1                                     if                                     y[i]                                     =                                     =                                     '1'                                     else                                     0                       
                                                 result                                     =                                     (                                     '1'                                     if                                     r                                     %                                     2                                     =                                     =                                     1                                     else                                     '0'                                     )                                     +                                     result                       
                                                 carry                                     =                                     0                                     if                                     r <                                     2                                     else                                     1                                     # Compute the carry.                       
                                   
                                                 if                                     carry !                                     =                                     0                                     : result                                     =                                     '1'                                     +                                     result                       
           
                                                 return                                     result.zfill(max_len)                       
           
                         # Driver code                       
                         print                                     (add_binary_nums(                                     '1101'                                     ,                                     '100'                                     ))                       
           
                         # This code is contributed                       
                         # by Anand Khatri                       

C#

                         // C# program to add                       
                         // two binary strings                       
                         using                                     System;                       
           
                         class                                     GFG {                       
                                   
                                                 // This function adds two                       
                                                 // binary strings and return                       
                                                 // result as a third string                       
                                                 static                                     string                                     addBinary(                                     string                                     a,                       
                                                 string                                     b)                       
                                                 {                       
                                   
                                                 // Initialize result                       
                                                 string                                     result =                                     ""                                     ;                       
                                   
                                                 // Initialize digit sum                       
                                                 int                                     s = 0;                       
           
                                                 // Traverse both strings starting                       
                                                 // from last characters                       
                                                 int                                     i = a.Length - 1, j = b.Length - 1;                       
                                                 while                                     (i >= 0 || j >= 0 || s == 1)                       
                                                 {                       
                                   
                                                 // Comput sum of last                       
                                                 // digits and carry                       
                                                 s += ((i >= 0)? a[i] -                                     '0'                                     : 0);                       
                                                 s += ((j >= 0)? b[j] -                                     '0'                                     : 0);                       
           
                                                 // If current digit sum is                       
                                                 // 1 or 3, add 1 to result                       
                                                 result = (                                     char                                     )(s % 2 +                                     '0'                                     ) + result;                       
           
                                                 // Compute carry                       
                                                 s /= 2;                       
           
                                                 // Move to next digits                       
                                                 i--; j--;                       
                                                 }                       
                                                 return                                     result;                       
                                                 }                       
                                   
                         // Driver Code                       
                         public                                     static                                     void                                     Main()                       
                         {                       
                                                 string                                     a =                                     "1101"                                     , b=                                     "100"                                     ;                       
                                                 Console.Write( addBinary(a, b));                       
                         }                       
                         }                       
           
                         // This code is contributed by Sam007                       

PHP

                         <?php                       
                         // PHP program to add two binary strings                       
           
                         // This function adds two binary strings                       
                         // and return result as a third string                       
                         function                                     addBinary(                                     $a                                     ,                                     $b                                     )                       
                         {                       
                                                 $result                                     =                                     ""                                     ;                                     // Initialize result                       
                                                 $s                                     = 0;                                     // Initialize digit sum                       
           
                                                 // Traverse both strings starting                       
                                                 // from last characters                       
                                                 $i                                     =                                     strlen                                     (                                     $a                                     ) - 1;                       
                                                 $j                                     =                                     strlen                                     (                                     $b                                     ) - 1;                       
                                                 while                                     (                                     $i                                     >= 0 ||                                     $j                                     >= 0 ||                                     $s                                     == 1)                       
                                                 {                       
                                                 // Comput sum of last digits and carry                       
                                                 $s                                     += ((                                     $i                                     >= 0)? ord(                                     $a                                     [                                     $i                                     ]) -                       
                                                 ord(                                     '0'                                     ): 0);                       
                                                 $s                                     += ((                                     $j                                     >= 0)? ord(                                     $b                                     [                                     $j                                     ]) -                       
                                                 ord(                                     '0'                                     ): 0);                       
           
                                                 // If current digit sum is 1 or 3,                       
                                                 // add 1 to result                       
                                                 $result                                     =                                     chr                                     (                                     $s                                     % 2 + ord(                                     '0'                                     )) .                                     $result                                     ;                       
           
                                                 // Compute carry                       
                                                 $s                                     = (int)(                                     $s                                     / 2);                       
           
                                                 // Move to next digits                       
                                                 $i                                     --;                                     $j                                     --;                       
                                                 }                       
                                                 return                                     $result                                     ;                       
                         }                       
           
                         // Driver Code                       
                         $a                                     =                                     "1101"                                     ;                       
                         $b                                     =                                     "100"                                     ;                       
                         echo                                     addBinary(                                     $a                                     ,                                     $b                                     );                       
           
                         // This code is contributed by mits                       
                         ?>                       

Javascript

                         <script>                       
           
                         // Javascript program to add                       
                         // two binary strings                       
           
                         // This function adds two                       
                         // binary strings and return                       
                         // result as a third string                       
                         function                                     addBinary(a, b)                       
                         {                       
                                   
                                                 // Initialize result                       
                                                 var                                     result =                                     ""                                     ;                       
                                   
                                                 // Initialize digit sum                       
                                                 var                                     s = 0;                       
           
                                                 // Traverse both strings starting                       
                                                 // from last characters                       
                                                 var                                     i = a.length - 1, j = b.length - 1;                       
                                                 while                                     (i >= 0 || j >= 0 || s == 1)                       
                                                 {                       
                                   
                                                 // Comput sum of last                       
                                                 // digits and carry                       
                                                 s += ((i >= 0)? a.charAt(i).charCodeAt(0) -                       
                                                 '0'                                     .charCodeAt(0): 0);                       
                                                 s += ((j >= 0)? b.charAt(j).charCodeAt(0) -                       
                                                 '0'                                     .charCodeAt(0): 0);                       
           
                                                 // If current digit sum is                       
                                                 // 1 or 3, add 1 to result                       
                                                 result = String.fromCharCode(parseInt(s % 2) +                       
                                                 '0'                                     .charCodeAt(0)) + result;                       
           
                                                 // Compute carry                       
                                                 s = parseInt(s/2);                       
           
                                                 // Move to next digits                       
                                                 i--; j--;                       
                                                 }                       
                                   
                         return                                     result;                       
                         }                       
           
                         //Driver code                       
                         var                                     a =                                     "1101"                                     , b=                                     "100"                                     ;                       
           
                         document.write(addBinary(a, b));                       
           
                         // This code is contributed by Amit Katiyar                       
           
                         </script>                       

输出：

感谢Gaurav Ahirwar提出上述解决方案。如果您发现任何不正确的地方，或者您想分享有关上述主题的更多信息，请写评论

文章版权归作者所有，未经允许请勿转载。

THE END

技术文章