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]
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规划的方法来解决。 1、dp[l][r][k]表示区间[l,r]内,以字符k+...
https://github.com/grandyang/leetcode/issues/730 类似题目: Longest Palindromic Subsequence Longest Palindromic Substring Palindromic Substrings 参考资料: https://leetcode.com/problems/count-different-palindromic-subsequences/ https://leetcode.com/problems/count-different-palindromic-subsequences/discuss/109509...
// 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…
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730. Count Different Palindromic Subsequences 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....
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
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/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规划的方法来解决。 1、dp[l][r][k]表示区间[l,r]内,以字符k+...
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 ...