[LeetCode] Implement strStr()

Well, the problem does not aim for an advanced algorithm like KMP but only a clean brute-force algorithm. So we can traverse all the possible starting points of haystack (from 0 tohaystack.length() - needle.length()) and see if the following characters in haystack match those of needle.

The code is as follows.

 1 class Solution {
 2 public: 
 3     int strStr(string haystack, string needle) {
 4         int m = haystack.length(), n = needle.length();
 5         if (!n) return 0;
 6         for (int i = 0; i < m - n + 1; i++) {
 7             int j = 0;
 8             for (; j < n; j++)
 9                 if (haystack[i + j] != needle[j])
10                     break;
11             if (j == n) return i;
12         }
13         return -1;
14     }
15 };

Of course, you may challenge yourself implementing the KMP algorithm for this problem.

KMP is a classic and yet notoriously hard-to-understand algorithm. However, I think the following two links give nice explanations. You may refer to them.

  1. KMP on jBoxer's blog;
  2. KMP on geeksforgeeks, with a well-commented C code.

I am sorry that I am still unable to give a personal explanation of the algorithm. I only read it from the two links above and mimic the code in the second link.

My accepted C++ code using KMP is as follows. Well, it also takes 4ms -_-

 1 class Solution {
 2 public:
 3     int strStr(string haystack, string needle) {
 4         int m = haystack.length(), n = needle.length();
 5         if (!n) return 0;
 6         vector<int> lps = kmpProcess(needle);
 7         for (int i = 0, j = 0; i < m; ) {
 8             if (haystack[i] == needle[j]) { 
 9                 i++;
10                 j++;
11             }
12             if (j == n) return i - j;
13             if (i < m && haystack[i] != needle[j]) {
14                 if (j) j = lps[j - 1];
15                 else i++;
16             }
17         }
18         return -1;
19     }
20 private:
21     vector<int> kmpProcess(string& needle) {
22         int n = needle.length();
23         vector<int> lps(n, 0);
24         for (int i = 1, len = 0; i < n; ) {
25             if (needle[i] == needle[len])
26                 lps[i++] = ++len;
27             else if (len) len = lps[len - 1];
28             else lps[i++] = 0;
29         }
30         return lps;
31     }
32 };

 

posted @ 2015-06-06 22:27  jianchao-li  阅读(297)  评论(0)  编辑  收藏  举报