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?

方法1：Two Pointers

  • i指向s中的字符，j指向t中的字符
  • 扫描t，遇到一个跟s[i]相同的字符, i++.
  • 如果i == s.length()， 说明在t中能找到s
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public boolean isSubsequence(String s, String t) {
    if (s.length() == 0)
        return true;
    int i = 0, j = 0;
    for (; j < t.length() && i < s.length(); j++) {
        if (t.charAt(j) == s.charAt(i)) {
            i++;
        }
    }
    return i == s.length();
}