Given a string s and a string t, check if s is subsequence of t.

You may assume that there is only lower case English letters in both s and t. t is potentially a very long (length ~= 500,000) string, and sis a short string (<=100).

A subsequence of a string is a new string which is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (ie, "ace" is a subsequence of "abcde" while "aec" is not).

Example 1:
s = "abc", t = "ahbgdc"

Return true.

Example 2:
s = "axc", t = "ahbgdc"

Return false.

Follow up:
If there are lots of incoming S, say S1, S2, ... , Sk where k >= 1B, and you want to check one by one to see if T has its subsequence. In this scenario, how would you change your code?

题意:

给定串串s和t, 判断s是否是t的subsequence

思路:

指针count用来scan串串s

指针i用来scan串串t

从左至右,若t当前的字符等于s当前字符,count++

若count等于s.length(),说明s中的每个字符都包含在串串t中,最终说明s是t的subsequence

代码:

 class Solution {
public boolean isSubsequence(String s, String t) {
if(s == null || s.length() == 0) return true;
int count = 0;
for(int i = 0; i<t.length() ; i++){
if(t.charAt(i) == s.charAt(count)){
count++;
if(count == s.length()){return true;}
}
}
return false;
}
}

[leetcode]392. Is Subsequence 验证子序列的更多相关文章

  1. LeetCode 392. Is Subsequence

    Given a string s and a string t, check if s is subsequence of t. You may assume that there is only l ...

  2. LeetCode 376. Wiggle Subsequence 摆动子序列

    原题 A sequence of numbers is called a wiggle sequence if the differences between successive numbers s ...

  3. 392 Is Subsequence 判断子序列

    给定字符串 s 和 t ,判断 s 是否为 t 的子序列.你可以认为 s 和 t 中仅包含英文小写字母.字符串 t 可能会很长(长度 ~= 500,000),而 s 是个短字符串(长度 <=10 ...

  4. [leetcode] 392. Is Subsequence (Medium)

    原题 判断子序列 /** * @param {string} s * @param {string} t * @return {boolean} */ var isSubsequence = func ...

  5. LeetCode 392. Is Subsequence 详解

    题目详情 给定字符串 s 和 t ,判断 s 是否为 t 的子序列. 你可以认为 s 和 t 中仅包含英文小写字母.字符串 t 可能会很长(长度 ~= 500,000),而 s 是个短字符串(长度 & ...

  6. 【LeetCode】392. Is Subsequence 解题报告(Python)

    [LeetCode]392. Is Subsequence 解题报告(Python) 标签: LeetCode 题目地址:https://leetcode.com/problems/is-subseq ...

  7. [LeetCode] 125. Valid Palindrome 验证回文字符串

    Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignori ...

  8. 392. Is Subsequence

    392. Is Subsequence 水题,先是判断长度,长度t比s小,返回false,然后从左到右扫描t,然后同时扫描s,如果相同,s的index就往后拉一个,如果s的index等于s长度,返回t ...

  9. LeetCode:递增的三元子序列【334】

    LeetCode:递增的三元子序列[334] 题目描述 给定一个未排序的数组,判断这个数组中是否存在长度为 3 的递增子序列. 数学表达式如下: 如果存在这样的 i, j, k,  且满足 0 ≤ i ...

随机推荐

  1. GameObject数组

    一.声明 GameObject[] rock = ]; 二.遍历 可以先声明一个GameObject的数组GameObject[], 然后把需要遍历的对象放进去,然后可以用下面两种方式遍历:1.for ...

  2. Software Scalability with MapReduce

      Software Scalability with MapReduce Craig Henderson First published online April 2010 The architec ...

  3. 小朋友学C语言(7)

    数组 一.数组简介 C 语言支持数组数据结构,它可以存储一个固定大小的相同类型元素的顺序集合.数组是用来存储一系列数据,但它往往被认为是一系列相同类型的变量. 数组的声明并不是声明一个个单独的变量,比 ...

  4. express无中间件的增删改查

    index.js const express = require("express");导入express框架 const data = require("./data& ...

  5. Web 跨域请求(OCRS) 前端解决方案

    1.同源策略如下: URL 说明 是否允许通信 http://www.a.com/a.jshttp://www.a.com/b.js 同一域名下 允许 http://www.a.com/lab/a.j ...

  6. 同步锁源码分析(一)AbstractQueuedSynchronizer原理

    文章转载自 AbstractQueuedSynchronizer的介绍和原理分析 建议去看一下原文的评论,会有不少收获. 简介 AbstractQueuedSynchronizer 提供了一个基于FI ...

  7. python中发布订阅和主从配置

    发布订阅 发布者不是计划发送消息给特定的接收者(订阅者),而是发布的消息分到不同的频道,不需要知道什么样的订阅者订阅 订阅者对一个或多个频道感兴趣,只需接收感兴趣的消息,不需要知道什么样的发布者发布的 ...

  8. YUM软件包额外扩展了解项

    5.YUM配置文件 yum的配置一般有两种方式: 一种是直接配置/etc目录下的yum.conf文件, 另外一种是在/etc/yum.repos.d目录下增加.repo文件 [root@xuliang ...

  9. Some facts about topological sort

    Definition: a topological sort of a DAG G is a sort such that for all edge (i,j) in G, i precedes j. ...

  10. Hbase简介安装配置

    HBase —— Hadoop Database的简称 ,hbase 是分布式,稀疏的,持久化的,多维有序映射,它基于行键rowkey,列键column key,时间戳timestamp建立索引.它是 ...