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

NewImage

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

发表评论