# 【LeetCode】 392. Is Subsequence

## Description
> Given a string **s** and a string **t**, check if **s** is subsequence of **t**.

> 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).

> 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](https://leetcode.com/pbrother/) for adding this problem and creating all test cases.

> Constraints:
> * `0 <= s.length <= 100`
> * `0 <= t.length <= 10^4`
> * Both strings consists only of lowercase characters.

> 給予一個字串**s**和一個字串**t**，檢查**s**是否為**t**的子序列。

> 一個字串的子序列是指一個新字串，它透過刪除原字串中的部分字元(可以不刪除)且不移動其餘字元的相對位置所得到(例如，`"ace"`是`"abcde"`的子序列而`"aec"`不是)。

> 進階：
> 如果有一堆輸入 S，分別為 S1, S2, ... , Sk 且 k >= 1B，你想要一個個檢查T是否為它的子序列。在這樣的情況下，你要怎麼更改你的程式碼？

> 工作人員：
> 特別感謝[@pbrother](https://leetcode.com/pbrother/)加入這個問題和提供全部的測試資料。

> 限制：
> * `0 <= s.length <= 100`
> * `0 <= t.length <= 10^4`
> * 兩個字元都只包含小寫字元。

## Example:

Example 1:

Input: s = "abc", t = "ahbgdc"
Output: true


Example 2:

Input: s = "axc", t = "ahbgdc"
Output: false

## Solution

* 用一個箭頭(int)指向較短字串`s`的開頭，然後用迴圈跑一次較長的字串`t`。
* 當兩者相同時，將`s`的箭頭往下移一個字。
* 當`t`跑完之後，檢查`s`的箭頭是否指到尾端的後一格。

### Code

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

###### tags: `LeetCode` `C++`