Leetcode-392. Is Subsequence

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本文介绍了一种检查字符串是否为另一个字符串子序列的有效算法。通过迭代比对的方式，该算法能在较短时间内确定字符串s是否为字符串t的子序列。特别适用于s较短而t可能非常长的情况。

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前言：为了后续的实习面试，开始疯狂刷题，非常欢迎志同道合的朋友一起交流。因为时间比较紧张，目前的规划是先过一遍，写出能想到的最优算法，第二遍再考虑最优或者较优的方法。如有错误欢迎指正。博主首发优快云，mcf171专栏。

博客链接：mcf171的博客

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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?

这个题目说白了就是在子串中找当前字符在目标子串中是否有存在，如果有，往后继续找。Your runtime beats 12.40% of java submissions.

public class Solution {
    public boolean isSubsequence(String s, String t) {
        int i = 0, j = 0;
        boolean flag = true;
        while(i < s.length() && j < t.length()){
            if(s.charAt(i) == t.charAt(j)){
                i ++; j ++;
            }else j ++;
            
        }
        if(i != s.length()) return false;
        
        return flag;
    }
}