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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…
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 b…
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)…
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…
A character is unique in string S if it occurs exactly once in it. For example, in string S = "LETTER", the only unique characters are "L" and "R". Let's define UNIQ(S) as the number of unique characters in string S. For exam…
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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…
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