• 396. Rotate Function


    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.
    
    public int MaxRotateFunction(int[] A) {
            int size = A.Count();
            int res = 0;
            for(int i = 0; i< size; i++)
            {
                int sum = 0;
                int k = i;
                for(int j = 0;j< size;j++)
                {
                    sum +=  A[j]*((j+k)%size);
                }
                res  = i==0? sum:Math.Max(res,sum);
                
            }
            return res;
        }

    DP

    public int MaxRotateFunction(int[] A) {
            int size = A.Count();
            if(size == 0) return 0;
            var f = new int[size];
            int sum = 0;
            int add = 0;
            for(int i = 0; i< size; i++)
            {
                 sum +=  A[i]*i;
                 add += A[i]; 
            }
            f[0] = sum;
            int res = sum;
            for(int i = 1; i< size; i++)
            {
                f[i] = f[i-1] + add - size*A[size-i];
                res = Math.Max(res,f[i]);
            }
            return res;
        }
     
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  • 原文地址:https://www.cnblogs.com/renyualbert/p/5870493.html
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