暴力解法（O(mn)）：
1 class Solution { 2 public: 3 /** 4 * Returns a index to the first occurrence of target in source, 5 * or -1 if target is not part of source. 6 * @param source string to be scanned. 7 * @param target string containing the sequence of characters to match. 8 */ 9 int strStr(const char *source, const char *target) { 10 // write your code here 11 if (!source || !target) return -1; 12 int m = strlen(source), n = strlen(target); 13 for (int i = 0; i < m - n + 1; i++) { 14 int j = 0; 15 for (; j < n; j++) 16 if (source[i + j] != target[j]) 17 break; 18 if (j == n) return i; 19 } 20 return -1; 21 } 22 };
KMP（O(m + n)）：
1 class Solution { 2 public: 3 /** 4 * Returns a index to the first occurrence of target in source, 5 * or -1 if target is not part of source. 6 * @param source string to be scanned. 7 * @param target string containing the sequence of characters to match. 8 */ 9 int strStr(const char *source, const char *target) { 10 // write your code here 11 if (!source || !target) return -1; 12 int m = strlen(source), n = strlen(target); 13 if (!n) return 0; 14 vector<int> lps = kmpProcess(target); 15 for (int i = 0, j = 0; i < m; ) { 16 if (source[i] == target[j]) { 17 i++; 18 j++; 19 } 20 if (j == n) return i - j; 21 if (i < m && source[i] != target[j]) { 22 if (j) j = lps[j - 1]; 23 else i++; 24 } 25 } 26 return -1; 27 } 28 private: 29 vector<int> kmpProcess(const char* target) { 30 int n = strlen(target); 31 vector<int> lps(n, 0); 32 for (int i = 1, len = 0; i < n; ) { 33 if (target[i] == target[len]) 34 lps[i++] = ++len; 35 else if (len) len = lps[len - 1]; 36 else lps[i++] = 0; 37 } 38 return lps; 39 } 40 };