[leetcode]Implement strStr()

28. Implement strStr()

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

字符串模式匹配：
1.brute force: O(M×N)

int strStr(string haystack, string needle) {    int N1 = haystack.length(), N2 = needle.length();    if (N2 == 0) return 0;    for (int i = 0; i <= N1 - N2; ++i)        for (int j = 0; j < N2 && haystack[i + j] == needle[j]; ++j)            if (j == N2 - 1) return i;    return -1;}

2.KMP algorithm O(N)
算法的细节可以看这三个链接：1 , 2 和 3。

class Solution {private:    int M, N;    void computeLPS(vector<int> &lps, const string &needle)    {        // lps stands for longest proper suffix        lps.resize(M, 0);        // len stands for previous longest propere suffix        for (int i = 1, len = 0; i < M;)        {            if (needle[len] == needle[i])                lps[i++] = ++len;            else            {                // take AAACAAAA and i = 7 for example                // the previous len = 3                // needle[3] != needle[7]                // thus lps[7] is the same as string                // AAAA, which is 3.                if (len > 0)                    len = lps[len - 1];                else                    lps[i++] = 0;            }        }    }public:    int strStr(string haystack, string needle)    {        N = haystack.length(), M = needle.size();        vector<int> lps;        computeLPS(lps, needle);        for (int i = 0, j = 0; i <= N - M + j;)        {            if (haystack[i] == needle[j])                ++i, ++j;            else            {                if (j == 0)                    ++i;                else                    j = lps[j - 1];            }            if (j == M) return i - j;        }        return -1;    }};

ABABA
当j = 3时，lps = 2，前AB，后AB。
当j = 4时，判定p[lps[j - 1]] == p[j]，因此lps = 3，前ABC，后ABC。

ABABC
当j = 4时，判定p[lps[j - 1]] ！= p[j]，此时取上一步的最长前串AB，如果在后面加上C，此时的lps为0，所以ABAB在后面加上C后，lps为0。

3.Rabin-Karp Algorithm O(M+N)
Here is a slide about this algorithm.

int strStr(string haystack, string needle){    int N = haystack.length(), M = needle.length();    if (M == 0) return 0;    int q = 335549;             // table size    int d = 256;                // radix    int h = 1, h1 = 0, h2 = 0;    // precompute q^(M - 1)    for (int j = 1; j < M; ++j)        h = (h * d) % q;    for (int j = 0; j < M; ++j)    {        h1 = (h1 * d + needle[j]) % q;        h2 = (h2 * d + haystack[j]) % q;    }    for (int i = 0; i <= N - M; ++i)    {        if (h2 == h1)        {            int j;            for (j = 0; j < M; ++j)                if (haystack[i + j] != needle[j])                    break;            if (j == M) return i;        }        if (i < N - M)        {            h2 = (h2 - haystack[i] * h) % q;            h2 = (h2 * d + haystack[i + M]) % q;            if (h2 < 0) h2 += q;        }    }    return -1;}