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:算出每个F[i],求出最大值maxn~每个F[i]可由前一步F[i-1]得到,F[i] = F[i – 1] + sum – len * A[len – i],其中sum为A[i]总和,len为A[i]长度~找到这个关系后就能求得F[i]的所有值~
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class Solution { public: int maxRotateFunction(vector<int>& A) { int len = A.size(), sum = 0; if(len == 0) return 0; vector<int> F(len); for(int i = 0; i < len; i++) { sum += A[i]; F[0] += i * A[i]; } int maxn = F[0]; for(int i = 1; i < len; i++) { F[i] = F[i - 1] + sum - len * A[len - i]; maxn = max(F[i], maxn); } return maxn; } }; |
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