leetcode 392. Is Subsequence

发表于 | 分类于

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?

  • solution with loop

    1
    2
    3
    4
    5
    6
    7
    8
    9
    10
    11
    12
    13
    14
    class Solution(object):
        def isSubsequence(self, s, t):
            """
            :type s: str
            :type t: str
            :rtype: bool
            """
            if len(s) == 0:return True
            count = 0
            for c in t:
                if c == s[count]:
                    count += 1
                if count == len(s):return True
            return False

  • solution with two points

    1
    2
    3
    4
    5
    6
    7
    8
    9
    10
    11
    class Solution:
        def isSubsequence(self, s, t):
            i = 0
            j = 0
            while i < len(s) and j < len(t):
                if s[i] == t[j]:
                    i = i + 1
                    j = j + 1
                else:
                    j = j + 1
            return i == len(s)

  • solution with dp，it’s slow o(n^2)

    1
    2
    3
    4
    5
    6
    7
    8
    9
    10
    11
    12
    13
    14
    15
    16
    17
    18
            int m = s.length(), n = t.length();
            if (m == 0) return true;
            int[][] dp = new int[m+1][n+1];
            for (int  i = 0; i <= m; i++) {
                dp[i] = new int[n+1];
            }
            for (int i = 1; i <= m; i++) {
                for (int j = 1; j <= n; j++) {
                    if (s.charAt(i-1) == t.charAt(j-1)) {
                        dp[i][j] = dp[i-1][j-1] + 1;
                    }
                    else{
                        dp[i][j] = Math.max(dp[i-1][j], dp[i][j-1]);
                    }
                }
            }
            return dp[m][n] == m;
    }