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.

这道题算比较简单的一种，我们可以用两个指针分别指向字符串s和t，然后如果字符相等，则i和j自增1，反之只有j自增1，最后看i是否等于s的长度，等于说明s已经遍历完了，而且字符都有在t中出现过，参见代码如下：

解法一：

class Solution {
public:
    bool isSubsequence(string s, string t) {
        if (s.empty()) return true;
        int i = 0, j = 0;
        while (i < s.size() && j < t.size()) {
            if (s[i] == t[j]) {
                ++i; ++j;
            } else {
                ++j;
            }
        }
        return i == s.size();
    }
};

下面这种写法稍稍简洁了一些，但是思路并没有什么不同，参见代码如下：

解法二：

class Solution {
public:
    bool isSubsequence(string s, string t) {
        if (s.empty()) return true;
        int i = 0, j = 0;
        while (i < s.size() && j < t.size()) {
            if (s[i] == t[j]) ++i;
            ++j;
        }
        return i == s.size();
    }
};