【Leetcode】396. Rotate Function
阿新 • • 發佈:2019-01-26
題目連結:
題目:
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.
思路:
啊 沒想到這次contest中 我遇到最難的題是這道。。。用F(k)=F(k-1)-(n-1)*end+(sum-end) + 0*end = F(k-1)+sum-n*end
畫了個圖:
演算法:
public int maxRotateFunction(int[] A) { int max=Integer.MIN_VALUE; int sum = 0; int pre = 0; for(int i=0;i<A.length;i++){ pre +=A[i]*i; sum+=A[i]; } max = Math.max(pre, max); int k=1; while(k<A.length){ int res = pre+sum-A.length*A[A.length-k]; max = Math.max(max, res); pre = res; k++; } return max; }