[Leetcode] Implement strStr()

Implement strStr().

Returns a pointer to the first occurrence of needle in haystack, or null if needle is not part of haystack.

KMP算法！

class Solution {
public:
    void getNext(vector<int> &next, string &needle) {
        int i = 0, j = -1;
        next[i] = j;
        while (i != needle.length()) {
            while (j != -1 && needle[i] != needle[j]) j = next[j];
            next[++i] = ++j;
        }
    }
    int strStr(string haystack, string needle) {
        if (haystack.empty()) return needle.empty() ? 0 : -1;
        if (needle.empty()) return 0;
        vector<int> next(needle.length() + 1);
        getNext(next, needle);
        int i = 0, j = 0;
        while (i != haystack.length()) {
            while (j != -1 && haystack[i] != needle[j]) j = next[j];
            ++i; ++j;
            if (j == needle.length()) return i - j;
        }
        return -1;
    }
};

上面的是通常用的KMP算法，但是算法是有一定缺陷的。比如我们的模式串  pattern =“AAAAB”，其中很容易得到next数组为01230。如果目标匹配串为 “AAAACAAAAB” ，大家可以模拟一下，A要回溯多次。就是说我们的next数组优化并不彻底。优化算法：next[i]表示匹配串在i处如果匹配失败下次移到的位置。下面是优化后的的求next数组的代码。虽然两种写求得next值不一样，但是kmp函数的写法是一样的。

class Solution {
public:
    void getNext(vector<int> &next, string &needle) {
        int i = 0, j = -1;
        next[i] = j;
        while (i < needle.length() - 1) {
            while (j != -1 && needle[i] != needle[j]) j = next[j];
            ++i; ++j;
            //特殊情况，这里即为优化之处。考虑下AAAAB, 防止4个A形成0123在匹配时多次迭代。
            if (needle[i] == needle[j]) next[i] = next[j]; 
            else next[i] = j;
        }
    }
    int strStr(string haystack, string needle) {
        if (haystack.empty()) return needle.empty() ? 0 : -1;
        if (needle.empty()) return 0;
        vector<int> next(needle.length() + 1);
        getNext(next, needle);
        int i = 0, j = 0;
        while (i != haystack.length()) {
            while (j != -1 && haystack[i] != needle[j]) j = next[j];
            ++i; ++j;
            if (j == needle.length()) return i - j;
        }
        return -1;
    }
};

下面不是KMP算法，但是代码特别简洁：

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