LeetCode #396 - Rotate Function - Easy

Problem

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.

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Algorithm

整理一下题意：给定一各整数数组A，长度为n。假设Bk是A经过轮转k个位置得到的数组，定义轮转函数F如下，

F(k) = 0 * Bk[0] + 1 * Bk[1] + ... + (n-1) * Bk[n-1]。

要求返回F(k)的最大值。

数学问题。对于每个F(k)，利用F(k-1)得到F(k)，其递推关系为

F[i]=F[i-1]-A[(n-i)%n]*(n-1)+(sum-A[(n-i)%n]);

注意当A的长度为0或1时，F(k)恒等于0。

代码如下。

class Solution {public:    int maxRotateFunction(vector<int>& A) {        int n=A.size();        if(n==0||n==1) return 0;        vector<int> F(n,0);        int sum=0;        for(int i=0;i<A.size();i++){            sum+=A[i];            F[0]+=A[i]*i;        }        int max=F[0];        for(int i=1;i<n;i++){            F[i]=F[i-1]-A[(n-i)%n]*(n-1)+(sum-A[(n-i)%n]);            if(F[i]>max) max= F[i];        }        return max;    }};