[LintCode]文字列検索

 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 };

 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 };