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 sis 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"

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?

題目的意思很明顯，判斷s是否是t的子序列，可以不連續但是順序要對。當s為空時，屬於t的子序列。代碼如下：

class Solution(object):
    def isSubsequence(self, s, t):
        """
        :type s: str
        :type t: str
        :rtype: bool
        """
        if len(si) == 0:
            return True
        si = 0
        ti = 0
        while ti < len(t):
            
 
if s[si] == t[ti]:
                si += 1
            if si == len(s):
                return True
            ti += 1
        return False

Is Subsequence