按字母顺序打印两个字符串的常用字符-yiteyi-C++库

给定两个字符串，打印所有常用字符词典编纂的顺序如果没有常用字母，请打印-1。所有字母都是小写。

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例如：

Input : string1 : geeksstring2 : forgeeksOutput : eegksExplanation: The letters that are common between the two strings are e(2 times), k(1 time) and s(1 time).Hence the lexicographical output is "eegks"Input : string1 : hhhhhellostring2 : gfghhmhOutput : hhh

推荐：请尝试你的方法 {IDE} 首先，在进入解决方案之前。

其思想是使用字符计数数组。 1）计算第一个和第二个字符串中从“a”到“z”的所有字符的出现次数。将这些计数存储在两个数组a1[]和a2[]中。 2）导线a1[]和a2[]（注：两者的尺寸均为26）。对于每个索引i，打印字符“a”+i的次数等于min（a1[i]，a2[i]）。

以下是上述步骤的实施情况。

C++

                         // C++ program to print common characters                       
                         // of two Strings in alphabetical order                       
                         #include<bits/stdc++.h>                       
                         using                                     namespace                                     std;                       
           
                         int                                     main()                       
                         {                       
                                                 string s1 =                                     "geeksforgeeks"                                     ;                       
                                                 string s2 =                                     "practiceforgeeks"                                     ;                       
                                   
                                                 // to store the count of                       
                                                 // letters in the first string                       
                                                 int                                     a1[26] = {0};                       
                                   
                                                 // to store the count of                       
                                                 // letters in the second string                       
                                                 int                                     a2[26] = {0};                       
                                                 int                                     i , j;                       
                                                 char                                     ch;                       
                                                 char                                     ch1 =                                     'a'                                     ;                       
                                                 int                                     k = (                                     int                                     )ch1, m;                       
                                   
                                                 // for each letter present, increment the count                       
                                                 for                                     (i = 0 ; i < s1.length() ; i++)                       
                                                 {                       
                                                 a1[(                                     int                                     )s1[i] - k]++;                       
                                                 }                       
                                                 for                                     (i = 0 ; i < s2.length() ; i++)                       
                                                 {                       
                                                 a2[(                                     int                                     )s2[i] - k]++;                       
                                                 }                       
           
                                                 for                                     (i = 0 ; i < 26 ; i++)                       
                                                 {                       
                                                 // the if condition guarantees that                       
                                                 // the element is common, that is,                       
                                                 // a1[i] and a2[i] are both non zero                       
                                                 // means that the letter has occurred                       
                                                 // at least once in both the strings                       
                                                 if                                     (a1[i] != 0 and a2[i] != 0)                       
                                                 {                       
                                                 // print the letter for a number                       
                                                 // of times that is the minimum                       
                                                 // of its count in s1 and s2                       
                                                 for                                     (j = 0 ; j < min(a1[i] , a2[i]) ; j++)                       
                                                 {                       
                                                 m = k + i;                       
                                                 ch = (                                     char                                     )(k + i);                       
                                                 cout << ch;                       
                                                 }                       
                                                 }                       
                                                 }                       
                                                 return                                     0;                       
                         }                       

JAVA

                         // Java program to print common characters                       
                         // of two Strings in alphabetical order                       
                         import                                     java.io.*;                       
                         import                                     java.util.*;                       
           
                         // Function to find similar characters                       
                         public                                     class                                     Simstrings                       
                         {                       
                                                 static                                     final                                     int                                     MAX_CHAR =                                     26                                     ;                       
           
                                                 static                                     void                                     printCommon(String s1, String s2)                       
                                                 {                       
                                                 // two arrays of length 26 to store occurrence                       
                                                 // of a letters alphabetically for each string                       
                                                 int                                     [] a1 =                                     new                                     int                                     [MAX_CHAR];                       
                                                 int                                     [] a2 =                                     new                                     int                                     [MAX_CHAR];                       
           
                                                 int                                     length1 = s1.length();                       
                                                 int                                     length2 = s2.length();                       
           
                                                 for                                     (                                     int                                     i =                                     0                                     ; i < length1 ; i++)                       
                                                 a1[s1.charAt(i) -                                     'a'                                     ] +=                                     1                                     ;                       
           
                                                 for                                     (                                     int                                     i =                                     0                                     ; i < length2 ; i++)                       
                                                 a2[s2.charAt(i) -                                     'a'                                     ] +=                                     1                                     ;                       
           
                                                 // If a common index is non-zero, it means                       
                                                 // that the letter corresponding to that                       
                                                 // index is common to both strings                       
                                                 for                                     (                                     int                                     i =                                     0                                     ; i < MAX_CHAR ; i++)                       
                                                 {                       
                                                 if                                     (a1[i] !=                                     0                                     && a2[i] !=                                     0                                     )                       
                                                 {                       
                                                 // Find the minimum of the occurrence                       
                                                 // of the character in both strings and print                       
                                                 // the letter that many number of times                       
                                                 for                                     (                                     int                                     j =                                     0                                     ; j < Math.min(a1[i], a2[i]) ; j++)                       
                                                 System.out.print(((                                     char                                     )(i +                                     'a'                                     )));                       
                                                 }                       
                                                 }                       
                                                 }                       
           
                                                 // Driver code                       
                                                 public                                     static                                     void                                     main(String[] args)                                     throws                                     IOException                       
                                                 {                       
                                                 String s1 =                                     "geeksforgeeks"                                     , s2 =                                     "practiceforgeeks"                                     ;                       
                                                 printCommon(s1, s2);                       
                                                 }                       
                         }                       

Python3

                         # Python3 program to print common characters                       
                         # of two Strings in alphabetical order                       
           
           
                         # Initializing size of array                       
                         MAX_CHAR                                     =                                     26                       
           
                         # Function to find similar characters                       
                         def                                     printCommon( s1, s2):                       
                                                 # two arrays of length 26 to store occurrence                       
                                                 # of a letters alphabetically for each string                       
                                                 a1                                     =                                     [                                     0                                     for                                     i                                     in                                     range                                     (MAX_CHAR)]                       
                                                 a2                                     =                                     [                                     0                                     for                                     i                                     in                                     range                                     (MAX_CHAR)]                       
           
                                                 length1                                     =                                     len                                     (s1)                       
                                                 length2                                     =                                     len                                     (s2)                       
           
                                                 for                                     i                                     in                                     range                                     (                                     0                                     ,length1):                       
                                                 a1[                                     ord                                     (s1[i])                                     -                                     ord                                     (                                     'a'                                     )]                                     +                                     =                                     1                       
           
                                                 for                                     i                                     in                                     range                                     (                                     0                                     ,length2):                       
                                                 a2[                                     ord                                     (s2[i])                                     -                                     ord                                     (                                     'a'                                     )]                                     +                                     =                                     1                       
           
                                                 # If a common index is non-zero, it means                       
                                                 # that the letter corresponding to that                       
                                                 # index is common to both strings                       
                                                 for                                     i                                     in                                     range                                     (                                     0                                     ,MAX_CHAR):                       
                                                 if                                     (a1[i] !                                     =                                     0                                     and                                     a2[i] !                                     =                                     0                                     ):                       
                                   
                                                 # Find the minimum of the occurrence                       
                                                 # of the character in both strings and print                       
                                                 # the letter that many number of times                       
                                                 for                                     j                                     in                                     range                                     (                                     0                                     ,                                     min                                     (a1[i],a2[i])):                       
                                                 ch                                     =                                     chr                                     (                                     ord                                     (                                     'a'                                     )                                     +                                     i)                       
                                                 print                                     (ch, end                                     =                                     '')                       
                                   
           
                         # Driver code                       
                         if                                     __name__                                     =                                     =                                     "__main__"                                     :                       
                                                 s1                                     =                                     "geeksforgeeks"                       
                                                 s2                                     =                                     "practiceforgeeks"                       
                                                 printCommon(s1, s2);                       
           
                         # This Code is contributed by Abhishek Sharma                       

C#

                         // C# program to print common characters                       
                         // of two Strings in alphabetical order                       
           
                         using                                     System;                       
                         // Function to find similar characters                       
                         public                                     class                                     Simstrings                       
                         {                       
                                                 static                                     int                                     MAX_CHAR = 26;                       
                                   
                                                 static                                     void                                     printCommon(                                     string                                     s1,                                     string                                     s2)                       
                                                 {                       
                                                 // two arrays of length 26 to store occurrence                       
                                                 // of a letters alphabetically for each string                       
                                                 int                                     [] a1 =                                     new                                     int                                     [MAX_CHAR];                       
                                                 int                                     [] a2 =                                     new                                     int                                     [MAX_CHAR];                       
                                   
                                                 int                                     length1 = s1.Length;                       
                                                 int                                     length2 = s2.Length;                       
                                   
                                                 for                                     (                                     int                                     i = 0 ; i < length1 ; i++)                       
                                                 a1[s1[i] -                                     'a'                                     ] += 1;                       
                                   
                                                 for                                     (                                     int                                     i = 0 ; i < length2 ; i++)                       
                                                 a2[s2[i] -                                     'a'                                     ] += 1;                       
                                   
                                                 // If a common index is non-zero, it means                       
                                                 // that the letter corresponding to that                       
                                                 // index is common to both strings                       
                                                 for                                     (                                     int                                     i = 0 ; i < MAX_CHAR ; i++)                       
                                                 {                       
                                                 if                                     (a1[i] != 0 && a2[i] != 0)                       
                                                 {                       
                                                 // Find the minimum of the occurrence                       
                                                 // of the character in both strings and print                       
                                                 // the letter that many number of times                       
                                                 for                                     (                                     int                                     j = 0 ; j < Math.Min(a1[i], a2[i]) ; j++)                       
                                                 Console.Write(((                                     char                                     )(i +                                     'a'                                     )));                       
                                                 }                       
                                                 }                       
                                                 }                       
                                   
                                                 // Driver code                       
                                                 public                                     static                                     void                                     Main()                       
                                                 {                       
                                                 string                                     s1 =                                     "geeksforgeeks"                                     , s2 =                                     "practiceforgeeks"                                     ;                       
                                                 printCommon(s1, s2);                       
                                                 }                       
                         }                       

Javascript

                         <script>                       
                         // Javascript program to print common characters                       
                         // of two Strings in alphabetical order                       
                                   
                                                 let MAX_CHAR = 26;                       
                                                 // Function to find similar characters                       
                                                 function                                     printCommon(s1,s2)                       
                                                 {                       
                                                 // two arrays of length 26 to store occurrence                       
                                                 // of a letters alphabetically for each string                       
                                                 let a1 =                                     new                                     Array(MAX_CHAR);                       
                                                 let a2 =                                     new                                     Array(MAX_CHAR);                       
                                                 for                                     (let i=0;i<MAX_CHAR;i++)                       
                                                 {                       
                                                 a1[i]=0;                       
                                                 a2[i]=0;                       
                                                 }                       
                                   
                                                 let length1 = s1.length;                       
                                                 let length2 = s2.length;                       
                                   
                                                 for                                     (let i = 0 ; i < length1 ; i++)                       
                                                 a1[s1[i].charCodeAt(0) -                                     'a'                                     .charCodeAt(0)] += 1;                       
                                   
                                                 for                                     (let i = 0 ; i < length2 ; i++)                       
                                                 a2[s2[i].charCodeAt(0) -                                     'a'                                     .charCodeAt(0)] += 1;                       
                                   
                                                 // If a common index is non-zero, it means                       
                                                 // that the letter corresponding to that                       
                                                 // index is common to both strings                       
                                                 for                                     (let i = 0 ; i < MAX_CHAR ; i++)                       
                                                 {                       
                                                 if                                     (a1[i] != 0 && a2[i] != 0)                       
                                                 {                       
                                                 // Find the minimum of the occurrence                       
                                                 // of the character in both strings and print                       
                                                 // the letter that many number of times                       
                                                 for                                     (let j = 0 ; j < Math.min(a1[i], a2[i]) ; j++)                       
                                                 document.write((String.fromCharCode(i +                                     'a'                                     .charCodeAt(0))));                       
                                                 }                       
                                                 }                       
                                                 }                       
                                   
                                                 // Driver code                       
                                                 let s1 =                                     "geeksforgeeks"                                     , s2 =                                     "practiceforgeeks"                                     ;                       
                                                 printCommon(s1, s2);                       
                                   
                                   
                                   
                                                 // This code is contributed by avanitrachhadiya2155                       
                         </script>                       

输出：

eeefgkors

时间复杂性： 如果我们考虑n=长度（较大的字符串），则该算法运行在 O（n） 复杂性

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