目錄

• 題目描述
• 題目大意
• 解題方法
• 動態規劃
• 日期

題目描述

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"

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
    ^^^
 

題目大意

求S中有多少個子序列等於T。

解題方法

動態規劃

這個題一看就是DP。向字串序列問題確實有很多都是用DP求解的。

設dp陣列dp[i][j]表示S的前j個字元是T的前i個字元的子序列的個數為dp[i][j]。

那麼有dp[0][*] == 1，因為這個情況下，只能使用s的空字串進行匹配t。

如果s[j - 1] == t[i - 1]，那麼，dp[i][j] = dp[i - 1][j - 1] + dp[i][j - 1]，原因是t的前j個字元可以由s的前[i - 1]個字元和t的前[j - 1]個匹配的同時最後一個字元匹配，加上s的前[j - 1]個字元和t的前[i]個字元匹配同時丟棄s的第[j]個字元。

如果s[j - 1] != t[i - 1]，那麼dp[i][j] = dp[i][j - 1]，因為只能是前面的匹配，最後一個字元不能匹配，所以丟棄了。

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 —— 週一又開始了