Leetcode 392.Is Subsequence(第十一週作業)

首先貼出原題

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.

可以發現是判斷s是否所有字母都出現在了t中，而且要按順序出現。所以我採用了循環遍歷的方式，每次匹配一個字母后就分割一次字符串，循環調用。

#include<iostream>
#include<string>
using namespace std;
class Solution {
public:
	bool isSubsequence(string s, string t) {
		//如果s爲空串，則無論如何都是true
		if (s == "")
			return true;
		for (int i = 0; i < s.size(); i++) {
			for (int j = 0; j < t.size(); j++) {
				if (s[i] == t[j]) {
					//只剩一個字母也匹配，所以返回true
					if (s.size() == 1)
						return true;
					//將字符串的第一個已經匹配的字母剔掉
					string news = s.substr(i + 1,s.size()- i - 1);
					string newt = t.substr(j + 1, t.size() - j - 1);
					//循環調用
					return isSubsequence(news,newt);
				}
			}
			//如果循環整個t，都沒有這個字母，則返回false
			return false;
		}
		//返回false
		return false;
	}
};

可以看出複雜度爲O(s.size()*t.size());

空間複雜度是O（t.size()）。