本文共 1175 字,大约阅读时间需要 3 分钟。
(1) null
(2) length is 0
(3) return value
(4) strlen
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 here11 if (source == NULL || target == NULL) return -1;12 13 int i, j, len_s = strlen(source), len_t = strlen(target);14 15 if (len_s == 0 && len_t == 0) return 0;16 17 i = 0;18 while (source[i] != '\0') {19 if (i + len_t > len_s) return -1;20 j = 0;21 while (target[j] != '\0') {22 if (source[i + j] == target[j]) {23 j++;24 } else {25 break;26 }27 }28 if (target[j] == '\0') return i;29 i++;30 }31 return -1;32 }33 };
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