zoukankan      html  css  js  c++  java
  • LeetCode 813. Largest Sum of Averages

    原题链接在这里:https://leetcode.com/problems/largest-sum-of-averages/

    题目:

    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 necessarily integers.

    Example:
    Input: 
    A = [9,1,2,3,9]
    K = 3
    Output: 20
    Explanation: 
    The best choice is to partition A into [9], [1, 2, 3], [9]. The answer is 9 + (1 + 2 + 3) / 3 + 9 = 20.
    We could have also partitioned A into [9, 1], [2], [3, 9], for example.
    That partition would lead to a score of 5 + 2 + 6 = 13, which is worse.

    Note:

    • 1 <= A.length <= 100.
    • 1 <= A[i] <= 10000.
    • 1 <= K <= A.length.
    • Answers within 10^-6 of the correct answer will be accepted as correct.

    题解:

    Let dp[k][i] denotes largest sum of averages of A up to index i in to k adjacent parts.

    For j from k-1 to i, separate them into 2 sections, first k-1 parts from k-1 to j, second 1 part from j to i.

    Then dp[k][i] = max(dp[k-1][j] + ave(j-i)).

    可以降维.

    Time Complexity: O(k*n^2). n = A.length.

    Space: O(k*n). 

    AC Java: 

     1 class Solution {
     2     public double largestSumOfAverages(int[] A, int K) {
     3         if(A == null || A.length == 0){
     4             return 0;
     5         }
     6         
     7         int n = A.length;
     8         double [] sum = new double[n+1];
     9         double [][] dp = new double[K+1][n+1];
    10         for(int i = 1; i<=n; i++){
    11             sum[i] = sum[i-1]+A[i-1];
    12             dp[1][i] = sum[i]/i;
    13         }
    14         
    15         for(int k = 2; k<=K; k++){
    16             for(int i = k; i<=n; i++){
    17                 for(int j = k-1; j<i; j++){
    18                     dp[k][i] = Math.max(dp[k][i], dp[k-1][j]+(sum[i]-sum[j])/(i-j));
    19                 }
    20             }
    21         }
    22         
    23         return dp[K][n];
    24     }
    25 }
  • 相关阅读:
    iOS开发--UILabel可以显示
    网络编程之IO模型——IO模型比较分析
    网络编程之IO模型——异步IO
    网络编程之IO模型——多路复用IO
    网络编程之IO模型——非阻塞IO
    网络编程之IO模型——阻塞IO
    Linux基本命令
    Linux界面介绍
    Linux系统目录介绍
    Linux的前世今生
  • 原文地址:https://www.cnblogs.com/Dylan-Java-NYC/p/11452314.html
Copyright © 2011-2022 走看看