交换相邻两数 如果只是交换相邻两数,那么最少交换次数为该序列的逆序数. 交换任意两数 数字的总个数减去循环节的个数?? A cycle is a set of elements, each of which is in the place of another. So in example sequences { 2, 1, 4, 3}, there are two cycles: {1, 2} and {3, 4}. 1 is in the place where 2 needs to G
Given a non-empty integer array of size n, find the minimum number of moves required to make all array elements equal, where a move is incrementing n - 1 elements by 1. Example: Input: [1,2,3] Output: 3 Explanation: Only three moves are needed (remem
给定一个长度为 n 的非空整数数组,找到让数组所有元素相等的最小移动次数.每次移动可以使 n - 1 个元素增加 1.示例:输入:[1,2,3]输出:3解释:只需要3次移动(注意每次移动会增加两个元素的值):[1,2,3] => [2,3,3] => [3,4,3] => [4,4,4]详见:https://leetcode.com/problems/minimum-moves-to-equal-array-elements/description/ C++: class S
453. 最小移动次数使数组元素相等 给定一个长度为 n 的非空整数数组,找到让数组所有元素相等的最小移动次数.每次移动可以使 n - 1 个元素增加 1. 示例: 输入: [1,2,3] 输出: 3 解释: 只需要3次移动(注意每次移动会增加两个元素的值): [1,2,3] => [2,3,3] => [3,4,3] => [4,4,4] class Solution { public int minMoves(int[] nums) { int moves = 0, min = In
We have two integer sequences A and B of the same non-zero length. We are allowed to swap elements A[i] and B[i]. Note that both elements are in the same index position in their respective sequences. At the end of some number of swaps, A and B are b
[试题描述]定义一个函数,输入一个有序数组生成最小高度二叉树 We will try to create a binary tree such that for each node, the number of nodes in the left subtree and the right subtree are equal, if possible Algorithm:1 Insert into the tree the middle element of the array2 Ins
题目描述 统计一个数字在排序数组中出现的次数 思路 最贱的方法依旧是count计数.. 当然,,看到有序数组就应该想到二分法,找到重复数字左边和右边的数字,然后两个相减就可以了 解答 方法1 count class Solution: def GetNumberOfK(self, data, k): # write code here if not data or len(data) ==0: return 0 return data.count(k) 方法2,不用count的计数方法 clas
