题目：
Implement strStr().
Returns the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack.
Update (2014-11-02):
The signature of the function had been updated to return the index instead of the pointer. If you still see your function signature returns a char * or String, please click the reload button to reset your code definition.
代码：
class Solution { public: int strStr(string haystack, string needle) { const size_t len_hay = haystack.size(); const size_t len_nee = needle.size(); if ( len_hay < len_nee ) return -1; bool if_match = true; for ( size_t i = 0; i <= len_hay-len_nee; ++i ) { if_match = true; for ( size_t j = 0; j < len_nee; ++j ){ if (haystack[i+j]!=needle[j]){ if_match = false; break; } } if ( if_match ){ return i; } } return -1; } };
Tips:
利用标志变量，暴力解决方法。时间复杂度O(m×n)，空间复杂度O(1)
貌似还有个KMP算法，这个可以做到O(m+n)的时间复杂度，但是空间复杂度牺牲到了O(m)。明天再看KMP算法，这个效率的提升还是蛮高的，直接提高到了线性复杂度。
原文：http://www.cnblogs.com/xbf9xbf/p/4474888.html