【LeetCode】730. Count Different Palindromic Subsequences 解题报告（Python）

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

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目录

• • 题目描述
  • 题目大意
  • 解题方法
  • • 记忆化搜索
    • 动态规划
  • 日期

题目地址：https://leetcode.com/problems/count-different-palindromic-subsequences/description/

题目描述

Given a string S, find the number of different non-empty palindromic subsequences in S, and return that number modulo 10^9 + 7.

A subsequence of a string S is obtained by deleting 0 or more characters from S.

A sequence is palindromic if it is equal to the sequence reversed.

Two sequences A_1, A_2, ... and B_1, B_2, ... are different if there is some i for which A_i != B_i.

Example 1:

Input:
S = 'bccb'
Output: 6
Explanation:
The 6 different non-empty palindromic subsequences are 'b', 'c', 'bb', 'cc', 'bcb', 'bccb'.
Note that 'bcb' is counted only once, even though it occurs twice.

Example 2:

Input:
S = 'abcdabcdabcdabcdabcdabcdabcdabcddcbadcbadcbadcbadcbadcbadcbadcba'
Output: 104860361
Explanation:
There are 3104860382 different non-empty palindromic subsequences, which is 104860361 modulo 10^9 + 7.

Note:

• The length of S will be in the range [1, 1000].
• Each character S[i] will be in the set {‘a’, ‘b’, ‘c’, ‘d’}.

题目大意

求一个字符串的有多少个回文子序列。

解题方法

记忆化搜索

这个题太难了，我也只是抄了花花酱的答案，花花有个40分钟的视频，讲得非常清楚，强烈大家看看。

class Solution:
    def countPalindromicSubsequences(self, S):
        """
        :type S: str
        :rtype: int
        """
        def count(S, i, j):
            if i > j: return 0
            if i == j: return 1
            if self.m_[i][j]:
                return self.m_[i][j]
            if S[i] == S[j]:
                ans = count(S, i + 1, j - 1) * 2
                l = i + 1
                r = j - 1
                while l <= r and S[l] != S[i]: l += 1
                while l <= r and S[r] != S[i]: r -= 1
                if l > r: ans += 2
                elif l == r: ans += 1
                else: ans -= count(S, l + 1, r - 1)
            else:
                ans = count(S, i + 1, j) + count(S, i, j - 1) - count(S, i + 1, j - 1)

            self.m_[i][j] = ans % (10 ** 9 + 7)
            return self.m_[i][j]

        n = len(S)
        self.m_ = [[None for _ in range(n)] for _ in range(n)]
        return count(S, 0, n - 1)

动态规划

待补

日期

2018 年 11 月 17 日 —— 美妙的周末，美丽的天气