https://leetcode.com/problems/rotate-function/description/
题目
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 10^5.
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.
题目翻译
题目解析
参考答案
#include <iostream>
#include <vector>
using namespace std;
class Solution {
public:
int maxRotateFunction(vector<int>& A) {
if (A.empty())
{
return 0;
}
int iMaxSum = 0;
int iPreSum = 0;
int iLastIndex = A.size() - 1;
int iMaxBase = A.size() - 1;
for (int i = 0; i < A.size(); i++)
{
iMaxSum = iMaxSum + i * A[i];
}
iPreSum = iMaxSum;
for (int i = 0; i < A.size() - 1; i++)
{
int iCurSum = iPreSum;
for (int j = 0; j < A.size(); j++)
{
if (j != iLastIndex)
{
iCurSum += A[j];
}
else
{
iCurSum -= A[j] * iMaxBase;
}
}
iLastIndex--;
if (iCurSum > iMaxSum)
{
iMaxSum = iCurSum;
}
iPreSum = iCurSum;
}
return iMaxSum;
}
};
int main()
{
vector<int> A;
A.push_back(4);
A.push_back(3);
A.push_back(2);
A.push_back(6);
Solution o;
cout << o.maxRotateFunction(A) << endl;
return 0;
}
