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

Example：

Input: haystack = "hello", needle = "ll"
Output: 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().

解释下题目：

实现java的indexOf()方法

1. 挨个判断

实际耗时：6ms

public int strStr(String haystack, String needle) {
        if (needle.equals("") || null == needle) {
            return 0;
        }

        //最多循环haystack.length() - needle.length()次
        for (int i = 0; i <= haystack.length() - needle.length(); i++) {
            int now = i; //记录目前的haystack中的索引
            int count = 0; //匹配上的个数
            for (int j = 0; j < needle.length(); j++) {
                if (haystack.charAt(now) != needle.charAt(j)) {
                    break;
                } else {
                    count++;
                    now++;
                }
                if (count == needle.length()) {
                    return i;
                }
            }
        }
        return -1;
    }

  思路：那除了挨个判断还能怎么做呢？

时间复杂度O(nm) n和m分别为两个数组的长度

空间复杂度O(1)