[Leetcode] Rotate Function

来源:互联网 发布:帝国cms栏目id修改 编辑:程序博客网 时间:2024/04/27 19:12

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 = 25F(1) = (0 * 6) + (1 * 4) + (2 * 3) + (3 * 2) = 0 + 4 + 6 + 6 = 16F(2) = (0 * 2) + (1 * 6) + (2 * 4) + (3 * 3) = 0 + 6 + 8 + 9 = 23F(3) = (0 * 3) + (1 * 2) + (2 * 6) + (3 * 4) = 0 + 2 + 12 + 12 = 26So the maximum value of F(0), F(1), F(2), F(3) is F(3) = 26.


public class Solution {    public int maxRotateFunction(int[] A) {        int sum1=0;        int sum2=0;        if(A.length<2) return 0;        for(int i=0;i<A.length;i++)          {              sum1+=A[i]*i;              sum2+=A[i];          }        int max=sum1;        int max2=sum1;         for(int i=0;i<A.length;i++)         {          max2+=sum2-A.length*A[A.length-1-i];             if(max2>max) max=max2;         }         return max;    }}


0 0