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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 = http://www.mamicode.com/0; 5 int sum = 0; 6 int maxValue =http://www.mamicode.com/ 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 = http://www.mamicode.com/value + (sum-A[j])-A[j]*(A.length-1); 15 maxValue =http://www.mamicode.com/ Math.max(maxValue, value); 16 } 17 return maxValue; 18 } 19 }
Leetcode: Rotate Function
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