一、问题描述：

https://leetcode.com/problems/implement-strstr/

Implement strStr().

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

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二、实现：

暴力匹配

    int strStr(string haystack, string needle) {
    	if (haystack.length() < needle.length()) {
    		return -1;
    	}
    	if (haystack.length() == needle.length()) {
    		if (haystack == needle) {
    			return 0;
    		}
    		return -1;
    	}
        for (int i = 0; i < haystack.length()-needle.length()+1; ++i) {
        	string tmp = haystack.substr(i, needle.length());
        	cout << tmp << endl;
        	if (tmp == needle) {
        		return i;
        	}
        }
        return -1;
    }

KMP

注意大坑！

j<(int)needle.length()

不加(int)的话，由于length()为无符号类型，当j==-1时将会被转化为INT_MAX！

public:
    int strStr(string haystack, string needle) {
    	if (needle.empty()) return 0;
    	if (haystack.empty()) return -1;
    	if (haystack.length() == needle.length()) {
    		if (haystack == needle) return 0;
    		return -1;
    	}
    	vector<int> nextArray(needle.length(), 0);
    	generateNext(nextArray, needle);
    	int i = 0, j =0;
    	while(i<haystack.length() && j<(int)needle.length()) {
    		if (j==-1 || haystack[i]==needle[j]) {
    			i++;
    			j++;
    		}
    		else j = nextArray[j];
    	}
    	if (j == needle.length()) return i-j;
    	return -1;
    }
private:
    void generateNext(vector<int> &nextArray, string needle) {
    	nextArray[0] = -1;
    	int k = -1, j = 0;
    	while(j < needle.length()-1) {
    		if (k==-1 || needle[k]==needle[j]) {
    			k++;
    			j++;
    			if (needle[j] != needle[k]) nextArray[j] = k;
    			else nextArray[j] = nextArray[k];
    		}
    		else k = nextArray[k];
    	}
    }

====================================================================================================================== 最近做一道题，用到上面这种KMP写法，但是却超时了。 理论上10^6的长度应该是不会超过1s的，翻了一下算法导论，将上述KMP算法更改如下：

#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <vector>
using namespace std;
void generateNext(vector<int> &nextArray, string needle) {
	nextArray[0] = -1;
	int k=-1, q;
	for (q = 1; q < needle.length(); ++q) {
		while (k>=0 && needle[k+1]!=needle[q])
			k = nextArray[k];
		if (needle[k+1] == needle[q]) 
			k++;
		nextArray[q] = k;
	}
	// for (int i = 0; i < needle.length(); ++i) {
	// 	cout << nextArray[i] << endl;
	// }
}
int match(string hayStack, string needle, vector<int>& nextArray) {
	if (hayStack.length() < needle.length()) return 0;
	if (hayStack.length() == needle.length()) {
		if (hayStack == needle) return 1;
		return 0;
	}
	int q=-1, i=0;
	int cnt=0;
	for (i = 0; i < hayStack.length(); ++i) {
		while (q>=0 && needle[q+1]!=hayStack[i])
			q = nextArray[q];
		if (needle[q+1] == hayStack[i])
			q++;
		if (q == needle.length()-1) {
			cnt++;
			q = nextArray[q];
		}
	}
	return cnt;
}
int main(int argc, char const *argv[]) {
	string hayStack, needle;
	while (cin >> hayStack >> needle) {
		vector<int> nextArray(needle.length(), 0);
		generateNext(nextArray, needle);
		int cnt = match(hayStack, needle, nextArray);
		cout << cnt << endl;
	}
	return 0;
}

附：KMP讲解 http://blog.csdn.net/v_july_v/article/details/7041827