leetcode 392. Is Subsequence 子序列判断 深度优先遍历DFS + 一个很简单的循环

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?

这道题是子序列的判定问题，最直接的想法就是DFS深度优先遍历来做，但是后来想一想有更加简单的方法，就是一个两层循环即可，递归做法会超时

代码如下：

#include <iostream>#include <vector>#include <map>#include <set>#include <queue>#include <stack>#include <string>#include <climits>#include <algorithm>#include <sstream>using namespace std;class Solution {public:    bool isSubsequence(string s, string t)     {        //return isSub(s, 0, t, 0);        int i = 0,j = 0;        while (i < t.length())        {            if (j == s.length())                return true;            else if (t[i] == s[j])            {                i++;                j++;            }            else                i++;        }         return j == s.length();    }    bool isSub(string s, int i, string t, int j)    {        if (i == s.length())            return true;        else        {            if (j == t.length())                return false;            else            {                if (s[i] == t[j])                    return isSub(s, i + 1, t, j + 1);                else                    return isSub(s, i, t, j + 1);            }        }    }};