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