【LeetCode】392. Is Subsequence

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本文介绍了一个简单的算法，用于检查一个字符串是否为另一个字符串的子序列。通过遍历长字符串并匹配短字符串中的字符来实现这一目标，该算法的时间复杂度为O(n)。

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题目：
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?
分析：
这道题要我们判断字符串s是否可以由字符串T删去某些字符构成。设变量len为当前在字符串T中找到的字符串s中字符的个数，遍历字符串T，当s[len]被找到后，len++,当len等于s的长度减1，即在字符串T中找到s，返回true。否则在遍历结束后返回false。时间复杂度为O(n).
代码：

class Solution {
public:
    bool isSubsequence(string s, string t) {
        if(t.length()<s.length())
        return false;
        if(s.length() == 0)
        return true;
        int len = 0;
        for(int i = 0;i<t.length();i++)
        {
            if(len == s.length()-1)
            return true;
            if(t[i] == s[len])
            len++;
        }
        return false;
    }
};