Given an integer array, return the k-th smallest distance among all the pairs. The distance of a pair (A, B) is defined as the absolute difference between A and B. Example 1: Input: nums = [1,3,1] k = 1 Output: 0 Explanation: Here are all the pairs:…

Given an integer array, return the k-th smallest distance among all the pairs. The distance of a pair (A, B) is defined as the absolute difference between A and B. Example 1: Input: nums = [1,3,1] k = 1 Output: 0 Explanation: Here are all the pairs:…

Given an integer array, return the k-th smallest distance among all the pairs. The distance of a pair (A, B) is defined as the absolute difference between A and B. Example 1: Input: nums = [1,3,1] k = 1 Output: 0 Explanation: Here are all the pairs:…

题目如下: 解题思路:对于这一类知道上限和下限,求第N位是什么的题目,可以先看看二分查找的方法可不可行.首先对nums进行排序,很显然任意两个元素距离绝对值最小是0,最大是nums[-1] - nums[0],所以第N小的距离肯定在 0 ~ (nums[-1] - nums[0]) 之间.采用二分查找的话,题目就变成了输入一个数值n,判断是不是第N小的距离.怎么判断n是否是第N小的距离呢?因为nums是有序的,对于nums[i]来说,只要找到(n + nums[i]) 在nums中所处的位置,设…

Given an integer array, return the k-th smallest distance among all the pairs. The distance of a pair (A, B) is defined as the absolute difference between A and B. Example 1: Input: nums = [1,3,1] k = 1 Output: 0 Explanation: Here are all the pairs:…

一.寻找两个有序数组的中位数 1.1 问题描述 给定两个大小为 m 和 n 的不同时为空的有序数组 nums1 和 nums2.找出这两个有序数组的中位数,并且要求算法的时间复杂度为 O(log(m + n)). 1.2 算法分析 题目要求的时间复杂度是 O(log(m + n)),要产生这样级别的时间复杂度只有采用二分查找法,用分治递归的思路来考虑这个问题. 需要转换题目中求中位数的问题为求第 k 小数的问题.如果 m + n 是奇数,那么寻找第 k = (m + n)/2 + 1 小的数即可…

Given an array of integers and an integer k, you need to find the number of unique k-diff pairs in the array. Here a k-diff pair is defined as an integer pair (i, j), where i and j are both numbers in the array and their absolute difference is k. Exa…

Nearly every one have used the Multiplication Table. But could you find out the k-th smallest number quickly from the multiplication table? Given the height m and the length n of a m * nMultiplication Table, and a positive integer k, you need to retu…

题目链接 链接:https://leetcode.com/problems/max-sum-of-rectangle-no-larger-than-k/description/ 题解&代码 1.暴力枚举所有的情况,时间复杂度O(n^2*m^2),实际耗时759 ms class Solution { public: int maxSumSubmatrix(vector<vector<int> >& matrix, int k) { int n=matrix.size…

719. 找出第 K 小的数对距离 这道题其实有那么一点二分猜答案的意思,也有很多类似题目,只不过这道题确实表达的不是很清晰不容易想到,题没问题,我的问题.既然是猜答案,那么二分边界自然就是距离最大值和距离最小值了,每次我们得到一个mid,**遍历数组(双指针或二分)**找到所有小于等于mid的数对数量cnt.遍历完之后如果cnt小于我们需要的k值,说明距离小了点,需要让左边界右移,这里其实也很好理解的.就想清楚一个问题,距离越大,那么包含的数对就越多,距离越小当然就越少了,当前mid距离包含的…