Leetcode 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.

这道题目主要有这三种方法。在Leetcode上并没有展现KMP和BM算法的优势，应该与OJ的测试样例有关；
暴力解法时间复杂度 O(MN), KMP 为 O(M+N), BM 为 O(N/M)

暴力解法：

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

KMP算法：

int strStr(string haystack, string needle) {        int nsize = needle.size();        int hsize = haystack.size();        if (nsize == 0) return 0;        int *table = new int[nsize];        memset(table, 0, sizeof(int)*nsize);        //building match table        for (int i = 1, j = 0; i < nsize - 1;){            if (needle[i] != needle[j]){                if (j>0){                    j = table[j - 1];                }                else{                    i++;                }            }            else{                table[i] = j + 1;                i++;                j++;            }        }        //matching        for (int i = 0, match_pos = 0; i < hsize;){            if (haystack[i] == needle[match_pos]){                if (match_pos == nsize - 1){                    return i - (nsize - 1);                }                else{                    i++;                    match_pos++;                }            }            else{                if (match_pos == 0){                    i++;                }                else{                    match_pos = table[match_pos - 1];                }            }        }        delete[]table;        return -1;    }

BM算法：

public class Solution {public String strStr(String haystack, String needle) {    if(haystack== null) return null;    if(needle==null || needle.length()==0) return haystack;    if(needle.length()>haystack.length()) return null;    int pat_length = needle.length();    int right[] = new int[256];    for(int i=0;i<256;i++)      right[i] =-1;    for(int i=0;i<pat_length;i++)      right[needle.charAt(i)] =i;    int rtn = search(right,haystack,needle);    if(rtn == haystack.length()) return null;    else         return haystack.substring(rtn);}public int search(int[] right, String haystack,String needle){    int M = haystack.length();    int N = needle.length();    int i,j;    int skip =0;    for(i=0; i<=M-N; i+=skip)    {        skip =0;        for(j=N-1;j>=0;j--)        {            if(needle.charAt(j)!=haystack.charAt(i+j))            {                skip = j-right[haystack.charAt(j+i)];                if(skip<=0) skip=1;                break;            }        }        if(skip ==0) return i;    }    return M;}