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  • Leetcode: Rotate Function

    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 }
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  • 原文地址:https://www.cnblogs.com/EdwardLiu/p/6120519.html
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