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...
(具体的转移代码中体现) const int mod=1e9+7; class Solution { public: int dfs(int l,int r,int k,string& S,vector<vector<vector<int> > >& dp) { if(r<l) return 0; if(l==r) return (k==S[l]-'a')?1:0; if(dp[l][r][k]>=0) return dp[l][r][k]; int& res=dp[...
Asubsequenceof an array is a sequence that can be derived from the array by deleting some or no elements without changing the order of the remaining elements. Two subsequences aredifferentif theset of indiceschosen are different. Example 1: Input:nums = [0,1,2,2]Output:3Explanation:The spe...
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classSolution {public:intcountPalindromicSubsequences(stringS) {intn =S.size(); vector<vector<int>> chars(26, vector<int>()); vector<vector<int>> memo(n +1, vector<int>(n +1,0));for(inti =0; i < n; ++i) { chars[S[i]-'a'].push_back(i); ...
[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...
My Solutions to Leetcode problems. All solutions support C++ language, some support Java and Python. Multiple solutions will be given by most problems. Enjoy:) 我的Leetcode解答。所有的问题都支持C++语言,一部分问题支持Java语言。近乎所有问题都会提供多个算
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 ...
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...