反正总是有人要赢，那为什么不能是我呢~

 [问题描述]

Implement strStr().

Returns a pointer to the first occurrence of needle in haystack, or null if needle is not part of haystack.

[解题思路]

1.KMP. 2.暴力

1 char *Solution::strStr(char* haystack, char* needle) 2 { 3     if (haystack == NULL || needle == NULL) 4         return NULL; 5     if (needle[0] == '\0') 6         return haystack; 7     int next[strlen(needle)];//计算nextval 8     int i = 0, j = -1; 9     next[i] = -1;10     while (needle[i] != '\0'){11         if (j == -1 || needle[i] == needle[j]){12             i++, j++; next[i] = j;13         }14         else{15             j = next[j];16         }17     }18     i = 0, j = 0;//匹配字符串19     char* ans = NULL;20     while (haystack[i] != '\0'){21         while (j >= 0 && haystack[i] != needle[j])22             j = next[j];23         i++; j++;24         if (needle[j] == '\0'){25             ans = haystack + i - j;26             return ans;27         }28     }29     return ans;30 }

暴力有效的方法1：

1 char *strStr(char *haystack, char *needle) 2  { 3     if(needle == NULL) return haystack; 4     else{ 5         int len1 =strlen(haystack),len2 = strlen(needle); 6         if(len2 == 0) return haystack; 7         for(int i = 0 ; i < len1-len2+1; ++ i){ 8             if(strncmp(haystack+i,needle,len2) == 0) return haystack+i; 9         }10         return NULL;11     }12 }

暴力有效的方法2：

1 char *strStr(char *haystack, char *needle)  2 { 3     int i,j;   4     for (i = j = 0; haystack[i] && needle[j];) {   5         if (haystack[i] == needle[j]) {   6             ++i;   7             ++j;   8         } else {   9             i = i - j + 1;  10             j = 0;  11         }  12     }  13     return needle[j] ? 0 : (haystack + i - j);  14 }