检查给定字符串是否为异形-yiteyi-C++库

给一根绳子 s .任务是检查给定字符串中的字符是否为异形。A. 异形图是一个单词、短语或句子，字母表中的任何字母都不会出现一次以上。

null

例如：

Input : S = "the big dwarf only jumps"Output : YesEach alphabet in the string S is occurredonly once.Input : S = "geeksforgeeks" Output : NoSince alphabet 'g', 'e', 'k', 's' occurredmore than once.

推荐：请尝试你的方法 {IDE} 首先，在进入解决方案之前。

其想法是创建一个大小为26的哈希数组，初始化为0。遍历给定字符串的每个字母表，如果第一次遇到该字母表，则在相应的哈希数组位置标记1，否则返回false。

以下是该方法的实施情况：

C++

                         // C++ Program to check whether the given string is Heterogram or not.                       
                         #include<bits/stdc++.h>                       
                         using                                     namespace                                     std;                       
           
                         bool                                     isHeterogram(                                     char                                     s[],                                     int                                     n)                       
                         {                       
                                                 int                                     hash[26] = { 0 };                       
                                   
                                                 // traversing the string.                       
                                                 for                                     (                                     int                                     i = 0; i < n; i++)                       
                                                 {                       
                                                 // ignore the space                       
                                                 if                                     (s[i] !=                                     ' '                                     )                       
                                                 {                       
                                                 // if already encountered                       
                                                 if                                     (hash[s[i] -                                     'a'                                     ] == 0)                       
                                                 hash[s[i] -                                     'a'                                     ] = 1;                       
                                   
                                                 // else return false.                       
                                                 else                       
                                                 return                                     false                                     ;                       
                                                 }                       
                                                 }                       
                                   
                                                 return                                     true                                     ;                       
                         }                       
           
                         // Driven Program                       
                         int                                     main()                       
                         {                       
                                                 char                                     s[] =                                     "the big dwarf only jumps"                                     ;                       
                                                 int                                     n =                                     strlen                                     (s);                       
                                                 (isHeterogram(s, n))?(cout <<                                     "YES"                                     ):(cout <<                                     "NO"                                     );                       
                                                 return                                     0;                       
                         }                       

JAVA

                         // Java Program to check whether the                       
                         // given string is Heterogram or not.                       
                         class                                     GFG {                       
                                   
                                                 static                                     boolean                                     isHeterogram(String s,                                     int                                     n)                       
                                                 {                       
                                                 int                                     hash[] =                                     new                                     int                                     [                                     26                                     ];                       
                                   
                                                 // traversing the string.                       
                                                 for                                     (                                     int                                     i =                                     0                                     ; i < n; i++)                       
                                                 {                       
                                                 // ignore the space                       
                                                 if                                     (s.charAt(i) !=                                     ' '                                     )                       
                                                 {                       
                                                 // if already encountered                       
                                                 if                                     (hash[s.charAt(i) -                                     'a'                                     ] ==                                     0                                     )                       
                                                 hash[s.charAt(i) -                                     'a'                                     ] =                                     1                                     ;                       
                                   
                                                 // else return false.                       
                                                 else                       
                                                 return                                     false                                     ;                       
                                                 }                       
                                                 }                       
                                   
                                                 return                                     true                                     ;                       
                                                 }                       
                                   
                         // Driver code                       
                         public                                     static                                     void                                     main (String[] args)                       
                         {                       
                                                 String s =                                     "the big dwarf only jumps"                                     ;                       
                                                 int                                     n = s.length();                       
                                   
                                                 if                                     (isHeterogram(s, n))                       
                                                 System.out.print(                                     "YES"                                     );                       
                                                 else                       
                                                 System.out.print(                                     "NO"                                     );                       
                         }                       
                         }                       
           
                         // This code is contributed by Anant Agarwal.                       

Python3

                         # Python3 code to check                       
                         # whether the given                       
                         # string is Heterogram                       
                         # or not.                       
           
                         def                                     isHeterogram(s, n):                       
                                                 hash                                     =                                     [                                     0                                     ]                                     *                                     26                       
                                   
                                                 # traversing the                       
                                                 # string.                       
                                                 for                                     i                                     in                                     range                                     (n):                       
                                   
                                                 # ignore the space                       
                                                 if                                     s[i] !                                     =                                     ' '                                     :                       
                                   
                                                 # if already                       
                                                 # encountered                       
                                                 if                                     hash                                     [                                     ord                                     (s[i])                                     -                                     ord                                     (                                     'a'                                     )]                                     =                                     =                                     0                                     :                       
                                                 hash                                     [                                     ord                                     (s[i])                                     -                                     ord                                     (                                     'a'                                     )]                                     =                                     1                       
                                   
                                                 # else return false.                       
                                                 else                                     :                       
                                                 return                                     False                       
                                   
                                                 return                                     True                       
           
                         # Driven Code                       
                         s                                     =                                     "the big dwarf only jumps"                       
                         n                                     =                                     len                                     (s)                       
           
                         print                                     (                                     "YES"                                     if                                     isHeterogram(s, n)                                     else                                     "NO"                                     )                       
           
                         # This code is contributed by "Sharad_Bhardwaj".                       

C#

                         // C# Program to check whether the                       
                         // given string is Heterogram or not.                       
                         using                                     System;                       
           
                         class                                     GFG {                       
                                   
                                                 static                                     bool                                     isHeterogram(                                     string                                     s,                                     int                                     n)                       
                                                 {                       
                                                 int                                     []hash =                                     new                                     int                                     [26];                       
                                   
                                                 // traversing the string.                       
                                                 for                                     (                                     int                                     i = 0; i < n; i++)                       
                                                 {                       
                                                 // ignore the space                       
                                                 if                                     (s[i] !=                                     ' '                                     )                       
                                                 {                       
                                                 // if already encountered                       
                                                 if                                     (hash[s[i] -                                     'a'                                     ] == 0)                       
                                                 hash[s[i] -                                     'a'                                     ] = 1;                       
                                   
                                                 // else return false.                       
                                                 else                       
                                                 return                                     false                                     ;                       
                                                 }                       
                                                 }                       
                                   
                                                 return                                     true                                     ;                       
                                                 }                       
                                   
                                                 // Driver code                       
                                                 public                                     static                                     void                                     Main ()                       
                                                 {                       
                                                 string                                     s =                                     "the big dwarf only jumps"                                     ;                       
                                                 int                                     n = s.Length;                       
                                   
                                                 if                                     (isHeterogram(s, n))                       
                                                 Console.WriteLine(                                     "YES"                                     );                       
                                                 else                       
                                                 Console.WriteLine(                                     "NO"                                     );                       
                                                 }                       
                         }                       
           
                         // This code is contributed by Vt_m.                       

PHP

                         <?php                       
                         // PHP Program to check                       
                         // whether the given string                       
                         // is Heterogram or not.                       
           
                         function                                     isHeterogram(                                     $s                                     ,                                     $n                                     )                       
                         {                       
                                                 $hash                                     =                                     array                                     ();                       
                                                 for                                     (                                     $i                                     = 0;                                     $i                                     < 26;                                     $i                                     ++)                       
                                                 $hash                                     [                                     $i                                     ] = 0;                       
                                   
                                                 // traversing the string.                       
                                                 for                                     (                                     $i                                     = 0;                                     $i                                     <                                     $n                                     ;                                     $i                                     ++)                       
                                                 {                       
                                                 // ignore the space                       
                                                 if                                     (                                     $s                                     [                                     $i                                     ] !=                                     ' '                                     )                       
                                                 {                       
                                                 // if already encountered                       
                                                 if                                     (                                     $hash                                     [ord(                                     $s                                     [                                     $i                                     ]) -                       
                                                 ord(                                     'a'                                     )] == 0)                       
                                                 $hash                                     [ord(                                     $s                                     [                                     $i                                     ]) -                       
                                                 ord(                                     'a'                                     )] = 1;                       
                                   
                                                 // else return false.                       
                                                 else                       
                                                 return                                     false;                       
                                                 }                       
                                                 }                       
                                                 return                                     true;                       
                         }                       
           
                         // Driven Code                       
                         $s                                     =                                     "the big dwarf only jumps"                                     ;                       
                         $n                                     =                                     strlen                                     (                                     $s                                     );                       
                         if                                     (isHeterogram(                                     $s                                     ,                                     $n                                     ))                       
                                                 echo                                     (                                     "YES"                                     );                       
                         else                       
                                                 echo                                     (                                     "NO"                                     );                       
                                   
                         // This code is contributed by                       
                         // Manish Shaw(manishshaw1)                       
                         ?>                       

Javascript

                         <script>                       
           
                         // Javascript program to check whether                       
                         // the given string is Heterogram or not.                       
                         function                                     isHeterogram(s, n)                       
                         {                       
                                                 var                                     hash = Array(26).fill(0);                       
                                   
                                                 // Traversing the string.                       
                                                 for                                     (                                     var                                     i = 0; i < n; i++)                       
                                                 {                       
                                   
                                                 // Ignore the space                       
                                                 if                                     (s[i] !=                                     ' '                                     )                       
                                                 {                       
                                   
                                                 // If already encountered                       
                                                 if                                     (hash[s[i].charCodeAt(0) -                       
                                                 'a'                                     .charCodeAt(0)] == 0)                       
                                                 hash[s[i].charCodeAt(0) -                       
                                                 'a'                                     .charCodeAt(0)] = 1;                       
                                   
                                                 // Else return false.                       
                                                 else                       
                                                 return                                     false                                     ;                       
                                                 }                       
                                                 }                       
                                                 return                                     true                                     ;                       
                         }                       
           
                         // Driver code                       
                         var                                     s =                                     "the big dwarf only jumps"                                     ;                       
                         var                                     n = s.length;                       
           
                         (isHeterogram(s, n)) ? (document.write(                                     "YES"                                     )) :                       
                                                 (document.write(                                     "NO"                                     ));                       
                                   
                         // This code is contributed by rutvik_56                       
           
                         </script>                       

输出：

YES

时间复杂性： O（N）

辅助空间： O（26）

文章版权归作者所有，未经允许请勿转载。

THE END

技术文章