396 Rotate Function
Given an array of integers A and letn to be its length.
Assume Bk to be an array obtained by rotating the arrayA 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.
題目解釋:對於一個給定的數組,對數組進行加權求和,權重就是元素所在的索引號,然後再對數組進行順時針旋轉,也就是所有元素向右移動一位,最後一位移到最前面。求輸出最大的和是多少。解題思路:一開始我想到的是用隊列存放權重,然後隊列元素分別與數組相乘求和,但是後來發現,這種做法簡直就是暴力太愚蠢。應該從中尋找規律,我們不妨設想,所有數組元素向右移動一位會和會發生什麼變化,結果我們可以看到,除了最後一位元素,所有元素的權重值都加了1,也就是和多加了A[0]+A[1]+...+A[N-2],而對於最後一位元素,被移到最前面,也就是A[n-1]的權重從(n-1)變成了0, 因此和減少了 (n-1)*A[n-1]
所以說,每一次移動的和S(new),和相對於上一次移動的和S(old)的關係如下:
S(new) = S(old) + { S - A[n-1] - (n-1)*A[n-1] } = S(old) + { S - n*A[n-1] }; // S = A[0]+A[2]+A[3]+...+A[n]
因此,每順時針移動一位,只需要進行一次計算就好。
class Solution {
public:
int maxRotateFunction(vector<int>& A) {
int n = A.size();
if (n == 0) return 0;
int s = 0, presum = 0;
for(int i = 0; i < n; i++){
s += A[i]; // S = A[0] + A[1] +...+A[n-1]
presum += i * A[i];
}
int MaxSum = presum;
for(int i = n-1; i >= 0; i--){
presum += s - n*A[i]; // S(new) = S(old) + { S - n*A[i] };
MaxSum = max(MaxSum, presum);
}
return MaxSum;
}
};