leetcode 396. Rotation function

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Ideas ： iteration
analysis ： Manually simulate , In fact, we can find the law
First , use sum Make a note of F(0), use k Record the cumulative sum
Simply simulate , Find the law
nums = [4,3,2,6]
F(0) = (0 * 4) + (1 * 3) + (2 * 2) + (3 * 6)
F(1) = (0 * 6) + (1 * 4) + (2 * 3) + (3 * 2)
F(2) = (0 * 2) + (1 * 6) + (2 * 4) + (3 * 3)
F(1) And F(0) comparison , You can see , One more. 4, One more. 3, One more. 2, Less 3 individual 6
F(2) And F(1) comparison , You can see , One more. 4, One more. 3, One more. 6, Less 3 individual 2

F(1) = F(0)+(k-nums[n-1])-(n-1)*nums[n-1]
F(2) = F(0)+(k-nums[n-2])-(n-1)nums[n-2]
Get the law ：F(x)=F(x-1)+k-nnums[n-x]; (1<=x<n)
Code ：

class Solution {
    
    public int maxRotateFunction(int[] nums) {
    
        int max = Integer.MIN_VALUE;
        int preSum = 0;
        int sum=0;
        int k=0;
        int len = nums.length;
        // initialization F(0) Sum cumulative sum 
        for(int i =0; i<len; i++){
    
            sum = sum + i * nums[i];
            k+=nums[i];
        }
        max = sum;
        for(int i=1;i<len;i++){
    
            preSum = sum;
            sum = preSum-len*nums[len-i]+k;
            max = Math.max(max,sum);
        }
        return max;
    }
}