一、 问题描述
Leecode第二十八题，题目为：
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().
问题理解为：
实现strStr ()。
返回haystack中needle的第一次出现的索引，如果needle不是haystack的一部分，返回-1。
例1:
输入:haystack = “hello”， needle = “ll”
输出:2
例2:
输入:haystack = “aaaaa”， needle = “bba”
输出:1
说明:
当needle是空字符串时，返回值? 这是一个非常适合在面试中问的问题。
对于这个问题，当指针为空字符串时，我们将返回0。这与C的strstr()和Java的indexOf()一致。
二、算法思路
1、
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三、实现代码
class Solution {
public:int strStr(string haystack, string needle) {if (needle.empty()) return 0;int x = haystack.size();int y = needle.size();if (x < y) return -1;for (int i = 0; i <= x - y; ++i){int j = 0;for (j = 0; j < y; ++j) {if (haystack[i + j] != needle[j])break;}if (j == y) return i;}return -1;}
};