Leetcode 392.Is Subsequence(第十一周作业)

首先贴出原题

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.

可以发现是判断s是否所有字母都出现在了t中，而且要按顺序出现。所以我采用了循环遍历的方式，每次匹配一个字母后就分割一次字符串，循环调用。

#include<iostream>#include<string>using namespace std;class Solution {public:bool isSubsequence(string s, string t) {//如果s为空串，则无论如何都是trueif (s == "")return true;for (int i = 0; i < s.size(); i++) {for (int j = 0; j < t.size(); j++) {if (s[i] == t[j]) {//只剩一个字母也匹配，所以返回trueif (s.size() == 1)return true;//将字符串的第一个已经匹配的字母剔掉string news = s.substr(i + 1,s.size()- i - 1);string newt = t.substr(j + 1, t.size() - j - 1);//循环调用return isSubsequence(news,newt);}}//如果循环整个t，都没有这个字母，则返回falsereturn false;}//返回falsereturn false;}};

可以看出复杂度为O(s.size()*t.size());

空间复杂度是O（t.size()）。