Given an array of integers A, consider all non-empty subsequences of A. For any sequence S, let the width of S be the difference between the maximum and minimum element of S. Return the sum of the widths of all subsequences of A.  As the answer may b…

Given an array of integers A, consider all non-empty subsequences of A. For any sequence S, let the width of S be the difference between the maximum and minimum element of S. Return the sum of the widths of all subsequences of A. As the answer may be…

题目如下: 解题思路:题目定义的子序列宽度是最大值和最小值的差,因此可以忽略中间值.首先对数组排序,对于数组中任意一个元素,都可以成为子序列中的最大值和最小值而存在.例如数组[1,2,3,4,5,6],对于元素3来说,由左边[1,2]组成的所有子序列都可以以3为最大值的,而右边[4,5,6]组成的所有子序列都可以以3为最小值.根据排列组合的原理:[1,2]可以组成的子序列个数为C(2,1) + C(2,2) ,而[4,5,6]可以组成的子序列个数为C(3,1) + C(3,2) + C(3,3)…

Given an array of integers A, consider all non-empty subsequences of A. For any sequence S, let the width of S be the difference between the maximum and minimum element of S. Return the sum of the widths of all subsequences of A. As the answer may be…

Given an array of integers A, consider all non-empty subsequences of A. For any sequence S, let the width of S be the difference between the maximum and minimum element of S. Return the sum of the widths of all subsequences of A. As the answer may be…

2019-10-14 17:00:10 问题描述: 问题求解: 如果暴力求解,时间复杂度是exponational的,因为这里是子序列而不是子数组.显然,直接枚举子序列是不太现实的了,那么可以怎么做呢? 切入点有两点: 1)数组的顺序对最后的结果是没有影响的,那么排序后的数组和原来的数组的结果是同样的,我们可以对原数组进行排序操作降低问题复杂性. 2)既然直接考虑序列的方案是不可行的,那么还有个思路就是去考虑每个数对最后结果的贡献.如果能想到这一点的话,其实本题就已经基本解决了,考虑到排序好的数…

All LeetCode Questions List(Part of Answers, still updating) 题目汇总及部分答案(持续更新中) Leetcode problems classified by company 题目按公司分类(Last updated: October 2, 2017) .   Top Interview Questions # Title Difficulty Acceptance 1 Two Sum Medium 17.70% 2 Add Two N…

# Title Solution Acceptance Difficulty Frequency     4 Median of Two Sorted Arrays       27.2% Hard     10 Regular Expression Matching       25.6% Hard     23 Merge k Sorted Lists       35.8% Hard     25 Reverse Nodes in k-Group       37.7% Hard    …

请点击页面左上角 -> Fork me on Github 或直接访问本项目Github地址:LeetCode Solution by Swift    说明:题目中含有$符号则为付费题目. 如:[Swift]LeetCode156.二叉树的上下颠倒 $ Binary Tree Upside Down 请下拉滚动条查看最新 Weekly Contest!!! Swift LeetCode 目录 | Catalog 序        号 题名Title 难度     Difficulty  两数之…

[2]Add Two Numbers (2018年12月23日,review) 链表的高精度加法. 题解:链表专题:https://www.cnblogs.com/zhangwanying/p/9797184.html [7]Reverse Integer (2018年12月23日, review) 给了一个32位的整数,返回它的reverse后的整数.如果reverse后的数超过了整数的范围,就返回 0. Example 1: Input: 123 Output: 321 Example 2:…