第一个只出现一次的字符

题目：在一个字符串中找到第一个只出现一次的字符。如输入abaccdeff，则输出b。

分析：这道题是2006年google的一道笔试题。

#include "stdio.h" 

char FirstNotRepeatingChar(char* pString)
{
    unsigned int i;
      // invalid input
      if(!pString)
            return 0;

      // get a hash table, and initialize it 
      const int tableSize = 256;
      unsigned int hashTable[tableSize];
      for(i = 0; i < tableSize; ++ i)
            hashTable[i] = 0;

      // get the how many times each char appears in the string
      char* pHashKey = pString;
      while(*(pHashKey) != '')
            hashTable[*(pHashKey++)] ++;

      // find the first char which appears only once in a string
      pHashKey = pString;
      while(*pHashKey != '')
      {
            if(hashTable[*pHashKey] == 1)
                  return *pHashKey;

            pHashKey++;
      }

      // if the string is empty 
      // or every char in the string appears at least twice
      return 0;
} 
 
 int main()
 { 
     printf("%c",FirstNotRepeatingChar("abaccdeff"));
     return 0;
 }

例二：字符串其实就是字符数组，上面的例一先来一个小例子

#include "stdio.h" 

char FirstNotRepeatingChar(char* pString)
{
    
      char* pHashKey = pString;
      while(*(pHashKey) != '')
      {
              printf("%c
",*(pHashKey++));
          
      }   
     // find the first char which appears only once in a string     
} 
 
 int main()
 { 
  FirstNotRepeatingChar("abaccdeff");
     return 0;
 }

等