LeetCode 28. Implement strStr()（实现子串定位）

原题网址：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.

方法一：穷举。

public class Solution {
    public int strStr(String haystack, String needle) {
        if (needle.length() == 0) return 0;
        if (haystack.length() < needle.length()) return -1;
        char[] ha = haystack.toCharArray();
        char[] na = needle.toCharArray();
        for(int i = 0; i < ha.length && i + na.length <= ha.length; i++) {
            boolean match = true;
            for(int j = 0; j < na.length; j++) {
                if (ha[i + j] != na[j]) {
                    match = false;
                    break;
                }
            }
            if (match) return i;
        }
        return -1;
    }
}

方法二：KMP算法。

public class Solution {
    public int strStr(String haystack, String needle) {
        char[] ha = haystack.toCharArray();
        char[] na = needle.toCharArray();
        int[] kmp = new int[na.length];
        for(int offset=Math.min(na.length-1, ha.length-na.length); offset>0; offset--) {
            for(int i=0; i+offset

另一种实现：

public class Solution {
    public int strStr(String haystack, String needle) {
        if (needle.length() == 0) return 0;
        if (haystack.length() < needle.length()) return -1;
        char[] ha = haystack.toCharArray();
        char[] na = needle.toCharArray();
        int[] kmp = new int[na.length];
        for(int i = Math.min(ha.length - na.length, na.length - 2); i >= 1; i--) {
            for(int j = i; j < na.length - 1; j++) {
                if (na[j - i] == na[j]) {
                    kmp[j + 1] = i;
                } else {
                    break;
                }
            }
        }

        int h = 0, n = 0;
        while (h + na.length <= ha.length) {
            if (ha[h + n] == na[n]) {
                n++;
                if (n == na.length) return h;
            } else if (kmp[n] == 0) {
                h += Math.max(n, 1);
                n = 0;
            } else {
                h += kmp[n];
                n -= kmp[n];
            }
        }
        return -1;
    }
}

有一篇很牛的KMP算法介绍，参考：http://blog.csdn.net/v_july_v/article/details/7041827

看了上文之后，重新写了一下，把kmp数组看成是最大公共前后缀的长度，这样理解非常方便！

public class Solution {
    public int strStr(String haystack, String needle) {
        if (needle.length() == 0) return 0;
        if (haystack.length() < needle.length()) return -1;
        
        char[] ha = haystack.toCharArray();
        char[] na = needle.toCharArray();
        int[] len = new int[na.length];

        for(int i = 1; i < na.length && i <= ha.length - na.length; i++) {
            int j = len[i - 1];
            while (j > 0 && na[j] != na[i]) j = len[j - 1];
            len[i] = j + (na[j]==na[i]? 1 : 0); 
        }

        int h = 0, n = 0;
        while (h + na.length <= ha.length) {
            if (ha[h + n] == na[n]) {
                n++;
                if (n == na.length) return h;
            } else if (n == 0) {
                h++;
            } else {
                h += n - len[n - 1];
                n = len[n - 1];
            }
        }
        return -1;
    }
}