The array [1,3,2] and [2,1,3] have exactly 1 inverse pair. Solution: This problem's dp solution is challenging to construct 1. reference the tutorial of the solution;https://leetcode.com/articles/k-inverse-pairs-array/ very great explanation for 8 approaches, I only came up with the...
Given two integersnandk, find how many different arrays consist of numbers from1tonsuch that there are exactlykinverse pairs. We define an inverse pair as following: Forithandjthelement in the array, ifi<janda[i]>a[j]then it's an inverse pair; Otherwise, it's not. Since the answer ma...
所以,为了提高大家的算法能力,这个公众号后续每天带大家做一道算法题,题目就从LeetCode上面选 ! 今天和大家聊的问题叫做K个逆序对数组,我们先来看题面: leetcode.cn/problems/k- For an integer array nums, an inverse pair is a pair of integers [i, j] where 0 <= i < j < nums.length and nums...
https://leetcode.com/problems/k-inverse-pairs-array/description/ 【题意】 给定n和k,求正好有k个逆序对的长度为n的序列有多少个,0<=k<=1000, 1<=n<=1000,答案模1e9+7 【思路】 dp[i][j]表示正好有j个逆序对的长度为i的序列的方案数,最终我们要求的就是dp[n][k] 考虑dp[i+1][j]和dp[i...
参考链接:[LeetCode] K Inverse Pairs Array K个翻转对数组 代码如下: #include <iostream> #include <vector> #include <map> #include <set> #include <queue> #include <stack> #include <string> #include <climits> #include <algorithm> #include <sstream> ...
The array [1,3,2] and [2,1,3] have exactly 1 inverse pair. Note: The integernis in the range [1, 1000] andkis in the range [0, 1000]. 给出两个整数n和k,找出所有包含从1到n的数字,且恰好拥有k个逆序对的不同的数组的个数。
lc 629. K Inverse Pairs Array https://leetcode.com/problems/k-inverse-pairs-array/ 更新:这个问题怎么都感觉有O(1)解法,解析解,真的规律太明显,Fibonacci sequence都有通项公式呢。 给一个set(1,2,3,4,5,6...n),输出有多少种排列使之有k个逆序对。 观察...
https://leetcode.com/problems/k-inverse-pairs-array/discuss/104825/Shared-my-C%2B%2B-O(n-*-k)-solution-with-explanation__EOF__ 本文作者:Veritas des Liberty 本文链接:https://www.cnblogs.com/h-hkai/p/10506927.html关于博主:评论和私信会在第一时间回复。或者直接私信我。版权声明:本博客所有...
题解:https://leetcode-cn.com/problems/k-inverse-pairs-array/solution/acmjin-pai-ti-jie-dong-tai-gui-hua-bian-qkb8z/ 代码改进:从TLE到AC 代码: TLE:时间复杂度O(n*n*k) classSolution {public:longlongdp[1005][1005];intkInversePairs(intn,intk) {//dp[i][j]:表示i个数字1~i,逆序对数...
class Solution { private static final int MOD = (int)1e9 + 7; public int kInversePairs(int n, int k) { int[][] cnt = new int[2][k+1]; cnt[0][0] = 1;// >0数字 0逆序对 -》 1组 cnt[1][0] = 1;// 1数字 0逆序对 -》 1组 // ni 几个数字,,ki 几组逆序对 for(...