LeetCode 392. Is Subsequence 题解

检查子序列算法

最新推荐文章于 2021-11-12 20:36:39 发布

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本文介绍了一个简单的O(n)算法来判断字符串s是否为字符串t的子序列。特别适用于s较短（<=100），而t可能非常长（长度约为500,000）的情况。文章提供了C++实现代码，并探讨了当面对大量输入字符串时如何优化解决方案。

392. Is Subsequence

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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?

题目分析：

如果s是t的子序列，那么j中会按s中字符的顺序依次包含s的每个字符。由此可得一O(n)算法。注意s为空字符串时的特殊情况。

代码展示：

class Solution {
public:
    bool isSubsequence(string s, string t) {
        int s1 = s.size();
        int s2 = t.size();
        if(!s1) return true;
        int j=0;
        int i=0;
        while(j<t.size())
        {
            if(t[j]==s[i])
            {
                i++;
                j++;
                if(i==s1) return true;
            }
            else
            {
                j++;
            }
        }
        return false;
    }
};