leetcode解题: Rotate Function (396)

发表于 | 分类于

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:

1
2
3
4
5
6
7
8
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.

解法1：找规律 O(N) Time + O(1) Space

这题用找规律的办法，找规律的时候把数字简化成A,B,C,D.
f(0) = 0A + 1B + 2C + 3D
f(1) = 0D + 1A + 2B + 3C
f(2) = OC + 1D + 2A + 3B

在考虑一个sum = 1A + 1B + 1C + 1D
那么可以得到
f(1) = f(0) + sum - 4D
f(2) = f(1) + sum - 4C
…
于是一个O(N)的解法就出来了
Java

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
public class Solution {
    public int maxRotateFunction(int[] A) {
        if (A == null || A.length == 0) {
            return 0;
        }
        int f = 0;
        int sum = 0;
        for (int i = 0; i < A.length; ++i) {
            f += i * A[i];
            sum += A[i];
        }
        
        int res = f; // f(0)
        for (int i = 1; i < A.length; ++i) {
            f = f + sum - A.length * A[A.length - i];
            res = Math.max(res, f);
        }
        
        return res;
        
    }
}