背景

来看一道leetcode题目：
Implement strStr().

Returns the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack.

让我们找到第一个子串的位置，这就是典型的字符串匹配问题。首先想到的就是暴力求解，时间复杂度O(mn).

BF算法实现

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

但是我们可以发现，其实有些字符串我们已经匹配过了，就不需要再次匹配了，这就是经典的KMP算法，时间复杂度为O(m + n)。KMP算法的讲的比较好的，可以参考这个博客：http://www.cnblogs.com/c-cloud/p/3224788.html

KMP算法实现

class Solution {
public:
    int strStr(string haystack, string needle) {
        if (haystack.size() == 0) {
            if (needle.size() == 0)
                return 0;
            else
                return -1;
        } else if (needle.size() == 0)
            return 0;
        if (haystack.size() < needle.size())
            return -1;

        vector<int> Next(needle.size());
        getNext(needle, Next);

        for (int i = 0; i < haystack.size(); ) {
            // 如果剩下的字符串比目标字符串更短，那么直接返回-1.
            if (haystack.size() - i < needle.size())
                return -1;
            int num = 0;
            int step = 1;
            for (int j = 0; j < needle.size(); j++) {
                if (haystack[i + j] == needle[j]) {
                    num++;
                } else
                    break;
            }
            if (num == needle.size())
                return i;
            if (num)
                step = num - Next[num - 1];
            i = i + step;
        }
        return -1;
    }
    void getNext(string needle, vector<int>& next) {
        int k = 0;
        next[0] = 0;
        for (int i = 1; i < needle.size(); i++) {
            while (k > 0 && needle[i] != needle[k])
                k = next[k - 1];
            if (needle[i] == needle[k])
                k++;
            next[i] = k;
        }
    }
};