LC 813. Largest Sum of Averages】的更多相关文章

We partition a row of numbers A into at most K adjacent (non-empty) groups, then our score is the sum of the average of each group. What is the largest score we can achieve? Note that our partition must use every number in A, and that scores are not…
[LeetCode]813. Largest Sum of Averages 解题报告(Python) 标签(空格分隔): LeetCode 作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 题目地址:https://leetcode.com/problems/largest-sum-of-averages/description/ 题目描述: We partition a row of numbers A into at most…
We partition a row of numbers A into at most K adjacent (non-empty) groups, then our score is the sum of the average of each group. What is the largest score we can achieve? Note that our partition must use every number in A, and that scores are not…
对于一个数组中的数进行分组,取每个组里的平均值进行加和的. 使用动态规划,其中dp[i][k]表示以i为结尾的有k个分组的,那么递推式为: dp[i][k]=dp[j][k-1]+(sum[i]-sum[j])/(i-j)的,那么当k=1的时候就初始化为组内的平均值的,其中j的初始化为k-2,因为是从0为起始点的. class Solution { public: double largestSumOfAverages(vector<int>& A, int K) { int n=A.…
题目如下: 解题思路:求最值的题目优先考虑是否可以用动态规划.记dp[i][j]表示在数组A的第j个元素后面加上第i+1 (i从0开始计数)个分隔符后可以得到的最大平均值,那么可以得到递归关系式: dp[i][j] = max(dp[i][j],dp[i-1][k]+float(sum(A[k+1:j+1]))/float(j-k)) . 代码如下: class Solution(object): def largestSumOfAverages(self, A, K): ""&quo…
We partition a row of numbers A into at most K adjacent (non-empty) groups, then our score is the sum of the average of each group. What is the largest score we can achieve? Note that our partition must use every number in A, and that scores are not…
We partition a row of numbers A into at most K adjacent (non-empty) groups, then our score is the sum of the average of each group. What is the largest score we can achieve? Note that our partition must use every number in A, and that scores are not…
""" We partition a row of numbers A into at most K adjacent (non-empty) groups, then our score is the sum of the average of each group. What is the largest score we can achieve? Note that our partition must use every number in A, and that s…
2018-07-12 23:21:53 问题描述: 问题求解: dp[i][j] : 以ai结尾的分j个部分得到的最大值 dp[i][j] = max{dp[k][j - 1] + (ak+1 + ... + ai) / (i - k)} k = [j - 2, i - 1] public double largestSumOfAverages(int[] A, int K) { double[][] dp = new double[A.length][K + 1]; int curSum =…
[抄题]: Given an array which consists of non-negative integers and an integer m, you can split the array into m non-empty continuous subarrays. Write an algorithm to minimize the largest sum among these m subarrays. Note:If n is the length of array, as…