【Leetcode】396. Rotate Function
Easy
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.
分析如下:
F(1)-F(0)=(1-0)*A[0]+(2-1)*A[1]+(3-2)*A[2]+(0-3)*A[3]
F(2)-F(1)=(2-1)*A[0]+(3-2)*A[1]+(0-3)*A[2]+(1-0)*A[3]
F(3)-F(2)=(3-2)*A[0]+(0-3)*A[1]+(1-0)*A[2]+(2-1)*A[3]
sum = A[0]+…+A[length-1]
F(n)-F(n-1)=-(length-1)*A[length-n]+sum-A[length-n]=-length*A[length-n]+sum
F(n)=F(n-1)-length*A[length-n]+sum
故:
题目链接:
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 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.思路:
啊 没想到这次contest中 我遇到最难的题是这道。。。用F(k)=F(k-1)-(n-1)*end+(sum-end) + 0*end = F(k-1)+sum-n*end
画了个图:
算法:
[java]
view plain copypublic 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;
}
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