LeetCode Distinct Subsequences
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LeetCode解题之Distinct Subsequences
原题
给定两个字符串S和T,求T有多少种从属于S的子序列的情况。或者说S可以删除它自己任意个字符,但是不能改变字符的相对位置,那一共有多少种删法可以使S变为T。
注意点:
- 删除任意个字符包括不删除字符
例子:
输入: s = “rabbbit”, t = “rabbit”
输出: 3
解题思路
典型的动态规划问题,dp[i][j]
表示字符串S[:i]和T[:j]的不同子序列数目,如果S[i-1]和T[j-1]不相等,那么只能在S[:i-1]和T[:j]中匹配,即dp[i][j] = dp[i-1][j];而当S[i-1]和T[j-1]相等时,可以是这两个字符正好匹配,也可以忽略S[i-1],使T[j-1]在S[:i-1]中匹配,所以dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j]
。进一步观察可以将二维数组压缩为一维,只需要从后往前计算即可。
AC源码
class Solution(object): def numDistinct(self, s, t): """ :type s: str :type t: str :rtype: int """ m = len(s) n = len(t) dp = [0 for __ in range(n + 1)] dp[0] = 1 for i in range(m): for j in range(n - 1, -1, -1): if t[j] == s[i]: dp[j + 1] += dp[j] return dp[-1]if __name__ == "__main__": assert Solution().numDistinct("rabbbit", "rabbit") == 3
欢迎查看我的Github (https://github.com/gavinfish/LeetCode-Python) 来获得相关源码。
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