1、题目描述

2.题目分析

字符串操作，注意边界条件即可。

3.代码

1  int strStr(string haystack, string needle) { 2         int n = needle.size(); 3         int res = -1; 4         if( needle.size() == 0) 5             return 0; 6         for(string::iterator it = haystack.begin() ; it != haystack.end(); ++it){ 7             if((*it) == needle[0]){ 8                 bool isEqual = true; 9                 bool isEnd = false;10                 for(int i = 0 ; i < n; i++){11                     if( it + i == haystack.end() ){12                         isEnd = true;13                         break;14                     }15                     if( needle[i] != *(it+i)){16                         isEqual = false;17                         break;18                     }19                 }20                 if( isEnd == true )21                     break;22                 if( isEqual == true){23                     res = it-haystack.begin();24                     break;25                 }26                 27             }28         }29         return res;30         31     }