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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.

解答

原理：

F(k) = 0 * Bk[0] + 1 * Bk[1] + ... + (n-1) * Bk[n-1]F(k-1) = 0 * Bk-1[0] + 1 * Bk-1[1] + ... + (n-1) * Bk-1[n-1]       = 0 * Bk[1] + 1 * Bk[2] + ... + (n-2) * Bk[n-1] + (n-1) * Bk[0]Then,F(k) - F(k-1) = Bk[1] + Bk[2] + ... + Bk[n-1] + (1-n)Bk[0]              = (Bk[0] + ... + Bk[n-1]) - nBk[0]              = sum - nBk[0]Thus,F(k) = F(k-1) + sum - nBk[0]What is Bk[0]?k = 0; B[0] = A[0];k = 1; B[0] = A[len-1];k = 2; B[0] = A[len-2];...

O(N^2）和O(N)

class Solution1 {
  //TLEpublic:    int maxRotateFunction(vector
   
    & A) {        if(A.size()==0) return 0;        vector
    
      F(A.size(),0);        for(int i=0;i
     
      res)                res=F[i];        }        return res;    }};class Solution {
  //O(N)public:    int maxRotateFunction(vector
      
       & A) {        int F=0,sum=0;        int len=A.size();        for(int i=0;i
       
        =0;i--)        {            F=F+sum-len*A[i];            maxF=max(F,maxF);        }        return maxF;    }};
       
      
     
    
   

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