396 Rotate Function 旋转函数】的更多相关文章

给定一个长度为 n 的整数数组 A .假设 Bk 是数组 A 顺时针旋转 k 个位置后的数组,我们定义 A 的“旋转函数” F 为:F(k) = 0 * Bk[0] + 1 * Bk[1] + ... + (n-1) * Bk[n-1].计算F(0), F(1), ..., F(n-1)中的最大值.注意:可以认为 n 的值小于 105.示例:A = [4, 3, 2, 6]F(0) = (0 * 4) + (1 * 3) + (2 * 2) + (3 * 6) = 0 + 3 + 4 + 18…
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]. Ca…
作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 题目地址:https://leetcode.com/problems/rotate-function/description/ 题目描述: 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 positio…
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]. Ca…
[抄题]: 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-…
#-*- coding: UTF-8 -*- #超时#        lenA=len(A)#        maxSum=[]#        count=0#        while count<lenA:#            tmpSum=0#            for i in xrange(lenA):#                tmpSum+=i*A[i-count]#            maxSum.append(tmpSum)#            coun…
一开始没察觉到0123 3012 2301 而不是 0123 1230 2301 的原因,所以也没找到规律,一怒之下brute-force.. public int maxRotateFunction(int[] A) { if(A.length == 0) return 0; int res = Integer.MIN_VALUE; for(int i = 0; i < A.length;i++) { int temp = 0; for(int j = i,total = 0; total <…
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 kpositions 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]. Cal…
396. 旋转函数 给定一个长度为 n 的整数数组 A . 假设 Bk 是数组 A 顺时针旋转 k 个位置后的数组,我们定义 A 的"旋转函数" F 为: F(k) = 0 * Bk[0] + 1 * Bk[1] + - + (n-1) * Bk[n-1]. 计算F(0), F(1), -, F(n-1)中的最大值. 注意: 可以认为 n 的值小于 105. 示例: A = [4, 3, 2, 6] F(0) = (0 * 4) + (1 * 3) + (2 * 2) + (3 * 6…
旋转函数 给定一个长度为 n 的整数数组 A . 假设 Bk 是数组 A 顺时针旋转 k 个位置后的数组,我们定义 A 的"旋转函数" F 为: F(k) = 0 * Bk[0] + 1 * Bk[1] + ... + (n-1) * Bk[n-1]. 计算F(0), F(1), ..., F(n-1)中的最大值. 注意:可以认为 n 的值小于 105. 示例: A = [4, 3, 2, 6] F(0) = (0 * 4) + (1 * 3) + (2 * 2) + (3 * 6)…