Leetcode: Rotate Function

最新推荐文章于 2025-07-10 23:03:39 发布

转载最新推荐文章于 2025-07-10 23:03:39 发布 · 81 阅读

本文介绍了一个关于数组旋转的问题，定义了一种特殊的“旋转函数”，并通过一个具体的例子详细解释了如何计算该函数在不同旋转状态下的最大值。文章还提供了一个Java实现的方法，展示了如何高效地求解此问题。

Given an array of integers A and let n to be its length.

Assume Bk to be an array obtained by rotating the array A k positions clock-wise, we define a "rotation function" F on A as follow:

F(k) = 0 * Bk[0] + 1 * Bk[1] + ... + (n-1) * Bk[n-1].

Calculate the maximum value of F(0), F(1), ..., F(n-1).

Note:
n is guaranteed to be less than 105.

Example:

A = [4, 3, 2, 6]

F(0) = (0 * 4) + (1 * 3) + (2 * 2) + (3 * 6) = 0 + 3 + 4 + 18 = 25
F(1) = (0 * 6) + (1 * 4) + (2 * 3) + (3 * 2) = 0 + 4 + 6 + 6 = 16
F(2) = (0 * 2) + (1 * 6) + (2 * 4) + (3 * 3) = 0 + 6 + 8 + 9 = 23
F(3) = (0 * 3) + (1 * 2) + (2 * 6) + (3 * 4) = 0 + 2 + 12 + 12 = 26

So the maximum value of F(0), F(1), F(2), F(3) is F(3) = 26.

F(0) = (0 * 4) + (1 * 3) + (2 * 2) + (3 * 6) = 0 + 3 + 4 + 18 = 25　　　　[4, 3, 2, 6]

F(1) = (0 * 6) + (1 * 4) + (2 * 3) + (3 * 2) = 0 + 4 + 6 + 6 = 16　　　　 [6, 4, 3, 2] diff = (4+3+2) - 6*(nums.length-1)

F(2) = (0 * 2) + (1 * 6) + (2 * 4) + (3 * 3) = 0 + 6 + 8 + 9 = 23　　 [2, 6, 4, 3]　　 diff = (6+4+3) - 2*(nums.length-1)

 1 public class Solution {
 2     public int maxRotateFunction(int[] A) {
 3         if (A==null || A.length==0) return 0;
 4         int value = 0;
 5         int sum = 0;
 6         int maxValue = Integer.MIN_VALUE;
 7         
 8         for (int i=0; i<A.length; i++) {
 9             value += i * A[i];
10             sum += A[i];
11         }
12         
13         for (int j=A.length-1; j>=0; j--) {
14             value = value + (sum-A[j])-A[j]*(A.length-1);
15             maxValue = Math.max(maxValue, value);
16         }
17         return maxValue;
18     }
19 }