1. Title Description

Rotation function

Given a length of n Array of integers for nums .

hypothesis arrk It's an array nums Clockwise rotation k Array after position , We define nums Of Rotation function F by ：

F(k) = 0 * arrk[0] + 1 * arrk[1] + … + (n - 1) * arrk[n - 1]
return F(0), F(1), …, F(n-1) Maximum of .

The generated test cases make the answers meet the requirements 32 position Integers .

Example 1:

 Input : nums = [4,3,2,6]
 Output : 26
 explain :
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
 therefore  F(0), F(1), F(2), F(3)  The maximum value in is  F(3) = 26 .

Example 2:

 Input : nums = [100]
 Output : 0

2. Ideas

2.1 Code

This question is very interesting , If you use a violent loop, you will be prompted to timeout , Therefore, the formula needs to be analyzed . Analyze with topic examples ,
Further observation and derivation , Adding an element in each step can get the sum of the whole array , namely

$$\begin{array}{rl}F(0)& =0\ast nums[0]+1\ast nums[1]+2\ast nums[2]+3\ast nums[3]\\ & \\ F(1)& =0\ast nums[3]+1\ast nums[0]+2\ast nums[1]+3\ast nums[2]\\ & =F(0)+nums[0]+nums[1]+nums[2]-3\ast nums[3]\\ & =F(0)+(nums[0]+nums[1]+nums[2]+nums[3])-4\ast nums[3]\\ & \\ F(2)& =0\ast nums[2]+1\ast nums[3]+2\ast nums[0]+3\ast nums[1]\\ & =F(1)+num[0]+num[1]+num[3]-3\ast num[2]\\ & =F(1)+(num[0]+nums[2]+num[1]+num[3])-4\ast num[2\\ & \\ F(3)& =0\ast nums[1]+1\ast nums[2]+2\ast nums[3]+3\ast nums[0]\\ & =F(2)+num[1]+num[2]+num[3]-3\ast num[0\\ & =F(2)+(nums[0]+num[1]+num[2]+num[3])-4\ast num[0]\end{array}$$

So we get the derivation formula , among i Is the angle mark to be subtracted from the formula
$$F_i(n)=F_i(n-1)+\sum _{i=0}^{n}nums-nums.Length\ast nums[i]$$

The code is as follows ：

class Solution {
    
    public int maxRotateFunction(int[] nums) {
    
    int sum = 0;
    int max = Integer.MIN_VALUE;
    int curr = 0;
    for (int i = 0; i < nums.length; i++) {
    
      sum += nums[i];
      curr += nums[i] * i;
    }
    max = Math.max(max,curr);
    for (int i = nums.length - 1; i > 0; i--) {
    
      curr += sum - nums.length * nums[i];
      max = Math.max(max,curr);
    }
    return max;
    }
}

2.2 test result

Pass the test

3. summary

• Violent cycle resolution will timeout
• Observe the example and deduce the formula