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

傷城~ 2022-06-04 05:25 170阅读 0赞

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);
                }
            }
        }
    
    };