leetcode 392. Is Subsequence

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?

对两个遍历，如果最后s全部遍历过则输出true。。。这样的问题是如果已出现 t 中没有的字符，那也会把 t 遍历一遍，浪费时间

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

2ms的做法，利用其 indexOf() 函数

public class Solution {
    public boolean isSubsequence(String s, String t) {
        if (s == null || t == null)
            return s == null && t == null;
        
        int tt = 0;
        for (int ss = 0; ss < s.length(); ss++) {
            int pos = t.indexOf(s.charAt(ss), tt);
            if (pos == -1) 
                return false;
            tt = pos + 1;
        }
        return true;
    }
}