LeetCode 28：Implement strStr()

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().

实现字符串查找，从头到尾遍历haystack每一个字符，如果needle.charAt(j) != haystack.charAt(currentIndex)，进入haystack下一个字符，继续匹配needle第一个字符

当needle.charAt(j) = haystack.charAt(currentIndex)，haystack和needle都移到下一个字符，检查是否一致，由下面的例子可以看出对于haystack每一个字符时，currentIndex = i + j，随i移动

因为只需要找到匹配的子串，如果有多个位置匹配上，只需要返回第一次的位置

package com.lihuia.leetcode28;

/**
 * Copyright (C), lihuia.com
 * FileName: Solution
 * Author:   lihui
 * Date:     2018/7/24
 */

public class Solution {
     public int strStr(String haystack, String needle) {
         int maxLen = haystack.length();
         int subLen = needle.length();
         
         int currentIndex = 0;
         int currentPos = 0;
         
         for (int i = 0; i < maxLen - subLen + 1; i++) {
             for (int j = 0; j < subLen; j++) {
                 currentIndex = i + j;
                 if (haystack.charAt(currentIndex) != needle.charAt(j)) {
                     break;
                 }
                 currentPos = j + 1;
             }
             
             if (currentPos == subLen) {
                 return i;
             }
         }
         return -1;
    }
}

遍历的字符数和子串长度一致，返回位置即可，这道题难度是Easy，感觉不太简单，还是太弱了