392 Is Subsequence

392. Is Subsequence

1. Question

Given a stringsand 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, andsis 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"

Returntrue.

Example 2: s="axc", t="ahbgdc"

Returnfalse.

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?

2. Implementation

(1) DP

思路: dp[i][j] 表示s[0...i - 1]是不是t[0...j - 1]的subsequence

class Solution {
    public boolean isSubsequence(String s, String t) {
        int m = s.length(), n = t.length();
        boolean[][] dp = new boolean[s.length() + 1][t.length() + 1];

        for (int i = 0; i <= m; i++) {
            for (int j = 0; j <= n; j++) {
                if (i == 0) {
                    dp[i][j] = true;
                }
                else if (j == 0) {
                    dp[i][j] = false;
                }
                else if (s.charAt(i - 1) == t.charAt(j - 1)) {
                    dp[i][j] = dp[i - 1][j - 1];
                }
                else {
                    dp[i][j] = dp[i][j - 1];
                }
            }
        }
        return dp[m][n];
    }
}

3. Time & Space Complexity

DP: 时间复杂度O(mn), 空间复杂度O(mn)