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

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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<na.length; i++) {                if (na[i] != na[i+offset]) {                    kmp[i+offset] = offset;                    break;                }            }        }        int h = 0, n = 0;        while (h + na.length <= ha.length && n < na.length) {            if (ha[h+n] == na[n]) {                n ++;            } else if (kmp[n] == 0) {                h += n + 1;                n = 0;            } else {                h += kmp[n];                n -= kmp[n];            }            if (n == na.length) return h;        }        return n == na.length? h : -1;    }}

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;    }}

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