392. Is Subsequence | LeetCode Dynamic Programming

最新推荐文章于 2023-03-15 16:30:54 发布

原创最新推荐文章于 2023-03-15 16:30:54 发布 · 375 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode #string

算法设计与原理练习题专栏收录该内容

29 篇文章

订阅专栏

本文介绍了一个简单的算法，用于判断一个较短的字符串是否为另一个较长字符串的子序列。通过逐字符对比的方式，该算法能够在较短时间内得出结论。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Description

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.
Difficulty: Medium

Thinking

本题要求判断短字符串是不是长字符串的子序列。方法很简单，从s和t的首位开始遍历两字符串进行比较，当s中的字符与t中字符相同时，s与t字符数组的下标都加1；不相同时，s下标固定，t的坐标加1，直到相同。当t字符串的遍历坐标大于t的长度，而s串的遍历下表仍小于s长度时，说明s不是t的子序列，函数返回false。否则返回true。需要注意的是，当s为空串时，s一定是t的子序列，函数返回true。

Solution

class Solution {
public:
    bool isSubsequence(string s, string t) {
        int slen = s.size();
        int tlen = t.size();
        int j = 0, i = 0;
        if(slen == 0) return true;
        for(; i < slen; i++){
            while(t[j] != s[i] && j < tlen) j++;
            if(j >= tlen) return false;
            j++;
        }
        return true;
    }
};