Implement .

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  and Java's .

 

用substring()来判断

注意：Java 7中，string.substring()的时间复杂度是O(n)，n为子串长度

time: O(mn), space: O(1)　　-- m: needle length, n: haystack length

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