[leetcode] Implement strStr()

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

mplement strStr().

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

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;    }};

KMP在上面方法中，会生成

  0 1  2  3  4  5 6  7          indice

  A B C D M A B D          V

  -1 0 0  0  0  0  1 2 0       next

next表示此处匹配失败后，j的取值；即下次j应该返回的值。实际是上次匹配成功的值，这次匹配要先++。

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()) {            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;    }};

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;    }};