[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.

class Solution {
public: 
    int strStr(string haystack, string needle) {
        int m = haystack.length(), n = needle.length();
        if (!n) return 0;
        for (int i = 0; i < m - n + 1; i++) {
            int j = 0;
            for (; j < n; j++)
                if (haystack[i + j] != needle[j])
                    break;
            if (j == n) return i;
        }
        return -1;
    }
};

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 -_-

class Solution {
public:
    int strStr(string haystack, string needle) {
        int m = haystack.length(), n = needle.length();
        if (!n) return 0;
        vector<int> lps = kmpProcess(needle);
        for (int i = 0, j = 0; i < m; ) {
            if (haystack[i] == needle[j]) { 
                i++;
                j++;
            }
            if (j == n) return i - j;
            if (i < m && haystack[i] != needle[j]) {
                if (j) j = lps[j - 1];
                else i++;
            }
        }
        return -1;
    }
private:
    vector<int> kmpProcess(string& needle) {
        int n = needle.length();
        vector<int> lps(n, 0);
        for (int i = 1, len = 0; i < n; ) {
            if (needle[i] == needle[len])
                lps[i++] = ++len;
            else if (len) len = lps[len - 1];
            else lps[i++] = 0;
        }
        return lps;
    }
};