作者： 负雪明烛
id： fuxuemingzhu
个人博客： http://fuxuemingzhu.cn/

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题目地址：https://leetcode.com/problems/distinct-subsequences/description/

题目描述

Given a string S and a string T, count the number of distinct subsequences of S which equals T.

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:

Input: S = "rabbbit", T = "rabbit"

Output: 3

Explanation:

As shown below, there are 3 ways you can generate "rabbit" from S.

(The caret symbol ^ means the chosen letters)

rabbbit

^^^^ ^^

rabbbit

^^ ^^^^

rabbbit

^^^ ^^^

Example 2:

Input: S = "babgbag", T = "bag"

Output: 5

Explanation:

As shown below, there are 5 ways you can generate "bag" from S.

(The caret symbol ^ means the chosen letters)

babgbag

^^ ^

babgbag

^^    ^

babgbag

^    ^^

babgbag

  ^  ^^

babgbag

    ^^^

题目大意

求S中有多少个子序列等于T。

解题方法

动态规划

这个题一看就是DP。向字符串序列问题确实有很多都是用DP求解的。

设dp数组dp[i][j]表示S的前j个字符是T的前i个字符的子序列的个数为dp[i][j]。

那么有dp[0][*] == 1，因为这个情况下，只能使用s的空字符串进行匹配t。

如果s[j - 1] == t[i - 1]，那么，dp[i][j] = dp[i - 1][j - 1] + dp[i][j - 1]，原因是t的前j个字符可以由s的前[i - 1]个字符和t的前[j - 1]个匹配的同时最后一个字符匹配，加上s的前[j - 1]个字符和t的前[i]个字符匹配同时丢弃s的第[j]个字符。

如果s[j - 1] != t[i - 1]，那么dp[i][j] = dp[i][j - 1]，因为只能是前面的匹配，最后一个字符不能匹配，所以丢弃了。

class Solution:

    def numDistinct(self, s, t):

        """

        :type s: str

        :type t: str

        :rtype: int

        """

        M, N = len(s), len(t)

        dp = [[0] * (M + 1) for _ in range(N + 1)]

        for j in range(M + 1):

            dp[0][j] = 1

        for i in range(1, N + 1):

            for j in range(1, M + 1):

                if s[j - 1] == t[i - 1]:

                    dp[i][j] = dp[i - 1][j - 1] + dp[i][j - 1]

                else:

                    dp[i][j] = dp[i][j - 1]

        return dp[-1][-1]

日期

2018 年 11 月 19 日 —— 周一又开始了