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?

Credits:

Special thanks to @pbrother for adding this problem and creating all test cases.
第一种方法，双指针直接遍历法，较之第二种方法慢。从前向后遍历t，遇到跟s指针下字符相等，s的指针向后移一位，当s的指针等于s的长度时，返回true。代码如下：
public class Solution {
public boolean isSubsequence(String s, String t) {
if (s.length() == 0) {
return true;
}
int indexS = 0, indexT = 0;
for (char ch: t.toCharArray()) {
if (ch == s.charAt(indexS)) {
indexS ++;
}
if (indexS == s.length()) {
return true;
}
}
return false;
}
}第二种方法，利用的是string的indexOf函数。遍历的是较短的s，根据s的各个字符串去t里面找，如果找不到当前的字符，立马返回false，如果可以顺利遍历完s，返回true。代码如下：
public class Solution {
public boolean isSubsequence(String s, String t) {
if (t.length() < s.length()) {
return false;
}
int prev = 0;
for (char ch: s.toCharArray()) {
prev = t.indexOf(ch, prev);
if (prev == -1) {
return false;
}
prev ++;
}
return true;
}
}