[LeetCode]题解（python）：115-Distinct Subsequences

题目来源：

　　https://leetcode.com/problems/distinct-subsequences/

题意分析：

　　给定字符串S和T，判断S中可以组成多少个T，T是S的子串。

题目思路：

　　这是一个动态规划的问题。设定ans[i][j]为s[:i] 组成t[:j]的个数。那么动态规划方程为，if s[i - 1] == t[j - 1]： ans[i][j] = ans[i - 1][j - 1] + ans[i - 1][j]；else： ans[i][j] = ans[i - 1][j]。

代码（python）：

class Solution(object):
    def numDistinct(self, s, t):
        """
        :type s: str
        :type t: str
        :rtype: int
        """
        m,n = len(s),len(t)
        ans = [[0 for i in range(n+1)] for i in range(m+1)]
        for i in range(m + 1):
            ans[i][0] = 1
        for i in range(1,m + 1):
            for j in range(1,n + 1):
                if s[i - 1] == t[j - 1]:
                    ans[i][j] = ans[i - 1][j - 1] + ans[i - 1][j]
                else:
                    ans[i][j] = ans[i-1][j]
        return ans[m][n]

时间： 2024-10-25 13:11:00

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