題目要求

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 = 26So the maximum value of F(0), F(1), F(2), F(3) is F(3) = 26.

Bk代表對數(shù)組A在位置k上進(jìn)行順時(shí)針的旋轉(zhuǎn)后生成的數(shù)組。F(k) = 0 * Bk[0] + 1 * Bk[1] + ... + (n-1) * Bk[n-1]，要求返回獲得的最大的F(k)的值。

暴力循環(huán)

按照題目的要求，執(zhí)行兩次循環(huán)即可以獲得F(k)的所有值，只需要從中比較最大值即可。

public int maxRotateFunction(int[] A) {if(A == null || A.length == 0) return 0;int max = Integer.MIN_VALUE;for(int i = 0 ; i < A.length ; i++) {int value = 0;for(int j = 0 ; i < A.length ; j++) {value += j * A[(j+i)%A.length];}max = Math.max(value, max);}return max;}

數(shù)學(xué)思路

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] F(k) = F(k-1) + sum - n*Bk[0]k = 0 Bk[0] = A[0] k = 1 Bk[0] = A[len-1] k = 2 Bk[0] = A[len-2] ... public int maxRotateFunction(int[] A) {if(A == null || A.length == 0) return 0;int F = 0;int sum = 0;for(int i = 0 ; i<A.length ; i++) {sum += A[i];F += i * A[i];}int max = F;for(int i = 1 ; i<A.length ; i++) {F += sum - A.length * A[A.length - i];max = Math.max(F, max);}return max;}

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