Question

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 s is 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?

Credits:

Special thanks to @pbrother for adding this problem and creating all test cases.

Solution

依次匹配就好了。

Code

class Solution {
public:
bool isSubsequence(string s, string t) {
if (s.length() > t.length())
return false;
if (s.length() == 0)
return true;
int index = 0;
for (int i = 0; i < t.length(); i++) {
if (t[i] == s[index])
index++;
}
if (index == s.length())
return true;
else
return false;
}
};

LeetCode——Is Subsequence的更多相关文章

  1. [LeetCode] Is Subsequence 是子序列

    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] Wiggle Subsequence 摆动子序列

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

  3. LeetCode "Wiggle Subsequence" !

    Another interesting DP. Lesson learnt: how you define state is crucial.. 1. if DP[i] is defined as, ...

  4. [LeetCode] Is Subsequence 题解

    前言 这道题的实现方法有很多,包括dp,贪心算法,二分搜索,普通实现等等. 题目 Given a string s and a string t, check if s is subsequence ...

  5. LeetCode "Is Subsequence"

    There are 3 possible approaches: DP, divide&conquer and greedy. And apparently, DP has O(n^2) co ...

  6. leetcode 376Wiggle Subsequence

    用dp解 1)up定义为nums[i-1] < nums[i] down nums[i-1] > nums[i] 两个dp数组, up[i],记录包含nums[i]且nums[i-1] & ...

  7. [LeetCode] Arithmetic Slices II - Subsequence 算数切片之二 - 子序列

    A sequence of numbers is called arithmetic if it consists of at least three elements and if the diff ...

  8. [LeetCode] Increasing Triplet Subsequence 递增的三元子序列

    Given an unsorted array return whether an increasing subsequence of length 3 exists or not in the ar ...

  9. [LeetCode] Longest Increasing Subsequence 最长递增子序列

    Given an unsorted array of integers, find the length of longest increasing subsequence. For example, ...

随机推荐

  1. 剖析Docker文件系统:Aufs与Devicemapper

    http://www.infoq.com/cn/articles/analysis-of-docker-file-system-aufs-and-devicemapper Docker镜像 典型的Li ...

  2. Yii2的主从数据库设置

    项目做大了,数据库主从还是不可少的.使用Yii框架开发,如何设置数据库的主从呢?其实很简单. 先说一个主数据库服务器和多个从数据库服务器的情况,修改配置文件 config/db.php ,其中 sla ...

  3. 剑指Offer——矩阵中的路径

    题目描述: 请设计一个函数,用来判断在一个矩阵中是否存在一条包含某字符串所有字符的路径.路径可以从矩阵中的任意一个格子开始,每一步可以在矩阵中向左,向右,向上,向下移动一个格子.如果一条路径经过了矩阵 ...

  4. 剑指Offer——复杂链表的复制

    题目描述: 输入一个复杂链表(每个节点中有节点值,以及两个指针,一个指向下一个节点,另一个特殊指针指向任意一个节点),返回结果为复制后复杂链表的head.(注意,输出结果中请不要返回参数中的节点引用, ...

  5. JXL导出Excel工具类

    将Excel中的数据读取到List<Map<String, Object>>集合中   package com.mvc.util;   import java.io.File; ...

  6. Android Studio 使用小技巧和快捷键

    Android Studio 使用小技巧和快捷键 Alt+回车 导入包,自己主动修正 Ctrl+N   查找类 Ctrl+Shift+N 查找文件 Ctrl+Alt+L  格式化代码 Ctrl+Alt ...

  7. TP自适应

    最近又要求职了,梳理了下这两年折腾的东西,发现有个产品很可惜,都开发完了,但是没上市.中兴的一款手表,我很喜欢那个金属壳子,结实,拿在手里沉甸甸,可以用来砸核桃. 当时调TP的时候,换了几个厂家,程序 ...

  8. XSS注入学习

    引贴: http://mp.weixin.qq.com/s?__biz=MzIyMDEzMTA2MQ==&mid=2651148212&idx=1&sn=cd4dfda0b92 ...

  9. linux内核源代码、配置与编译

    内核源代码下载:www.kernel.org Linux内核源代码采用树形结构进行组织,非常合理地把功能相关的文件都放在同一个子目录下,使得程序更具可读性. linux内核代码最好不要在windows ...

  10. json & pickle数据序列化

    序列化:把内存中的数据对象变成字符串 info = { 'name':'tom', 'age':22 } f = open("test.txt","w") f. ...