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?

給定字符串s和t，判斷s是否是t的子序列。
這里假定s和t都只含有小寫字母，t可以是非常長的字符串（長度約為500,000），s是短字符串（長度小于100）。
注意子序列是由原生序列中的某些元素按照相對位置不變的狀態(tài)組成的新序列。
進(jìn)階：若輸入的待檢測短序列s有很多個(gè)，而你又想一個(gè)一個(gè)檢測是否為t的子序列，如何改寫你的代碼？

思路
簡單的做法，t中依次對s的首個(gè)字母進(jìn)行匹配，成功則匹配s的下一個(gè)字母，直到s被匹配完成（返回true）或t掃描結(jié)束且s仍未匹配完（返回false），這里注意，對于s為空的情況，總是匹配為true。

class Solution {
public:
    bool isSubsequence(string s, string t) {
        if(s.size()>t.size()) return false;
        if(s=="")   return true;
        for(auto sit=s.begin(),tit=t.begin(); tit!=t.end();tit++){
            if(*sit==*tit)  sit++;
            if(sit==s.end())    return true;
        }
        return false;
    }
};