• 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.
    
    Anlysis:
    roateValue[i-1] = rotateValue[i] - (len-1)*A[i] + (sum-A[i])

    Solution:

    public class Solution {
        public int maxRotateFunction(int[] A) {
            if (A.length==0) return 0;
            if (A.length==1) return 0;
            
            int rotateValue = 0;
            int sum = 0;
            for (int i=0;i<A.length;i++){
                sum += A[i];
                rotateValue += i*A[i];
            }
            
            int maxValue = rotateValue;
            for (int i=A.length-1;i>=1;i--){
                rotateValue = rotateValue - (A.length-1)*A[i] + (sum-A[i]);
                maxValue = Math.max(maxValue,rotateValue);
            }
            return maxValue;
        }
    }
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  • 原文地址:https://www.cnblogs.com/lishiblog/p/5867296.html
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