LeetCode 28. Implement strStr()

原题链接在这里：https://leetcode.com/problems/implement-strstr/

题目:

Implement strStr().

Return the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack.

Example 1:

Input: haystack = "hello", needle = "ll"
Output: 2

Example 2:

Input: haystack = "aaaaa", needle = "bba"
Output: -1

Clarification:

What should we return when needle is an empty string? This is a great question to ask during an interview.

For the purpose of this problem, we will return 0 when needle is an empty string. This is consistent to C's strstr() and Java's indexOf().

题解：

Brute force的算法是从haystack头开始走，每次取后面长度与needle长度相等的substring, 看这个substring与needle是否相同即可.

If needle is empty, return 0.

Note: s.substring(i), i could be s.length(), and return "". Thus here, it is allowed to use i <= m - n.

Time Complexity: O(m*n), m = haystack.length(), n = needle.length().

Space: O(1).　　　 

AC Java:

class Solution {
    public int strStr(String haystack, String needle) {
        if(haystack == null || needle == null){
            return -1;
        }
        
        if(needle.length() == 0){
            return 0;
        }
        
        int m = haystack.length();
        int n = needle.length();
        
        for(int i = 0; i <= m - n; i++){
            if(haystack.substring(i, i + n).equals(needle)){
                return i;
            }
        }
        
        return -1;
    }
}

本题还可以采用KMP算法，这篇帖子说的很详细。

注意在求next数组时，while loop的条件是q<len-1, 因为下面还用到q++后的next[q]. next array的含义是到当前char的前一位的最大公共前缀后缀.

Time Complexity: O(m+n). m = haystack.length(), n = needle.length(). Space: O(n)

AC Java: 

public class Solution {
    public int strStr(String haystack, String needle) {
        if(haystack == null || needle == null || haystack.length() < needle.length()){
            return -1;
        }
        
        if(needle.length() == 0){
            return 0;
        }
        
        int i = 0;
        int j = 0;
        int len1 = haystack.length();
        int len2 = needle.length();
        int [] next = new int[len2];
        getNext(needle, next);
        while(i<len1 && j<len2){
            if(j == -1 || haystack.charAt(i) == needle.charAt(j)){
                i++;
                j++;
            }else{
                j = next[j];
            }
        }
        
        if(j == len2){
            return i-j;
        }else{
            return -1;
        }
    }
    
    private void getNext(String s, int [] next){
        int len = s.length();
        next[0] = -1;
        int p = -1;
        int q = 0;
        while(q < len-1){
            //s.charAt(p)表示前缀. s.charAt(q)表示后缀
            if(p == -1 || s.charAt(p) == s.charAt(q)){
                p++;
                q++;
                next[q] = p;
            }else{
                p = next[p];
            }
        }
    }
}

跟上Shortest Palindrome.

参见如下两篇文章：

http://blog.csdn.net/v_july_v/article/details/7041827

http://blog.csdn.net/Evan123mg/article/details/46834827