解法2 (https://leetcode.com/problems/count-different-palindromic-subsequences/discuss/109507/Java-96ms-DP-Solution-with-Detailed-Explanation) 求S[s..e] 的回文串数 当S[s] != S[e] 时,可得公式 dp[s][e] = dp[s+1][e] + dp[s][e-1] - dp[s+1][e-1]; 即(包含s + 两边都不含...
管理 leetcode 730 Count Different Palindromic Subsequences 题目链接: https://leetcode.com/problems/count-different-palindromic-subsequences/description/ 730.Count Different Palindromic Subsequences 题意 给你一个只包含a、b、c、d字符的字符串,总共有多少个不重复的回文子序列。 题解 容易想到这题可以用动态...
// LeetCode 2020 hard #707 // 730. Count Different Palindromic Subsequences // https://leetcode.com/problems/count-different-palindromic-subsequences/ // Runtime: 108 ms, faster than 87.87% of C++ o…
https://discuss.leetcode.com/topic/111483/java-96ms-dp-solution-with-detailed-explanation 本文转自博客园Grandyang的博客,原文链接:[LeetCode] Count Different Palindromic Subsequences 计数不同的回文子序列的个数 ,如需转载请自行联系原博主。
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花花酱 LeetCode 730. Count Different Palindromic Subsequences - 刷题找工作 EP114 思路上有很多细节要处理,但如果每一种情况都理清了的话也不难。 所有的情况: 主要的两大类是首位字符相同或不相同,首位字符相同的大类又分为三小类:就是除开首位两字符之外的中间字符含不含有首尾字符,不含?含有一个,还是含...
https://leetcode.com/problems/count-different-palindromic-subsequences/discuss/109509/Accepted-Java-Solution-using-memoization https://leetcode.com/problems/count-different-palindromic-subsequences/discuss/109507/Java-96ms-DP-Solution-with-Detailed-Explanation ...
原题链接在这里:https://leetcode.com/problems/count-different-palindromic-subsequences/ 题目: 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 ...
[leetcode]730. Count Different Palindromic Subsequences计数不同的回文子序列的个数 Example 1: Input: S= 'bccb'Output:6Explanation: The6 different non-empty palindromic subsequences are 'b', 'c', 'bb', 'cc', 'bcb', 'bccb'. Note that'bcb' is counted only once, even though it occurs...
1classSolution {2func countPalindromicSubsequences(_ S: String) ->Int {3vararr:[Character] =Array(S)4varn:Int =S.count5varM:Int = Int(1e9 +7)6vardp:[[Int]] = [[Int]](repeating:[Int](repeating:0,count:n),count:n)7foriin0..<n8{9dp[i][i] =110}11forlenin1..<n12{13for...