LeetCode:115. Distinct Subsequences

题目

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
^^^

解题思路

看了大佬们的解题思路，其实有点复杂，有点慢。

新建一个res[t_len+1, s_len+1]的全0矩阵，并令第一行为1
当t[i-1]==s[j-1]的时候，res[i][j] = res[i][j-1] + res[i-1][j-1]
否则 res[i][j] = res[i][j-1]

我第一条写得特别不美观！嫌弃！！！

class Solution:def numDistinct(self, s: str, t: str) -> int:l1, l2 = len(s)+1, len(t)+1res = []for i in range(len(t)+1):tmp = []for j in range(len(s)+1):tmp.append(0)res.append(tmp)for j in range(len(s)+1):res[0][j]=1for i in range(1,len(t)+1):for j in range(1, len(s)+1):print(t[i-1], s[j-1])if t[i-1]==s[j-1]:res[i][j] = res[i][j-1] + res[i-1][j-1] else:res[i][j] = res[i][j-1]return res[l2-1][l1-1]

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LeetCode:115. Distinct Subsequences

题目

解题思路

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