【LeetCode】028.Implement strStr()

题目：

Implement strStr().

Returns the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack.
解答：
这是典型的字符串匹配问题，用KMP算法（不清楚的请自行百度），KMP算法的核心就是利用最长前缀串减少回溯次数。

OJ上这道题有个坑：patStr长度为0时，必须返回0，不论tarStr是否为空。


代码如下：

import java.util.Vector;
public class Solution {
public int strStr(String haystack, String needle) {
if(haystack == null || needle == null)
return -1;
if(needle.length() == 0)
return 0;
if(haystack.length() == 0)
return -1;
char[] tar = haystack.toCharArray();
char[] pattern = needle.toCharArray();
Vector<Integer> prefix = this.CptPrefix(pattern);
int NOCM = 0; // Number of characters matched
for (int i = 0; i < tar.length; i++) {
while (NOCM > 0 && tar[i] != pattern[NOCM])
NOCM = prefix.elementAt(NOCM);
if (tar[i] == pattern[NOCM])
NOCM++;
if (NOCM == pattern.length) { // matched
return i - pattern.length + 1;
}
}
return -1;
}

public Vector<Integer> CptPrefix(char[] c) {
if (c == null || c.length == 0)
return null;
int NOCM = 0; // Number of Character Matched
int LOLP = 0; // Length of Longest Patter
Vector<Integer> prefix = new Vector<Integer>(c.length);
prefix.add(0);
prefix.add(0);
for (NOCM = 2; NOCM < c.length; NOCM++) {
while (LOLP > 0 && c[NOCM-1] != c[LOLP])
LOLP = prefix.elementAt(LOLP);
if (c[NOCM-1] == c[LOLP])
LOLP++;
prefix.add(NOCM, LOLP);
}
return prefix;
}
}