一道非常经典的题目,Median of Two Sorted Arrays.(PS:leetcode 我已经做了 190 道,欢迎围观全部题解 https://github.com/hanzichi/leetcode) 题意非常简单,给定两个有序的数组,求中位数,难度系数给的是 Hard,希望的复杂度是 log 级别.回顾下中位数,对于一个有序数组,如果数组长度是奇数,那么中位数就是中间那个值,如果长度是偶数,就是中间两个数的平均数. O(nlogn) 最容易想到的解法是 O(nlogn) 的解…

LeetCode(3) || Median of Two Sorted Arrays 题记 之前做了3题,感觉难度一般,没想到突然来了这道比较难的,星期六花了一天的时间才做完,可见以前基础太差了. 题目内容 There are two sorted arrays A and B of size m and n respectively. Find the median of the two sorted arrays. The overall run time complexity should…

题目来源:https://leetcode.com/problems/median-of-two-sorted-arrays/ There are two sorted arrays nums1 and nums2 of size m and n respectively. Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)). 解题思路: 题目是这样的:给…

4. Median of Two Sorted Arrays 题目链接:https://leetcode.com/problems/median-of-two-sorted-arrays/ Description: There are two sorted arrays nums1 and nums2 of size m and n respectively. Find the median of the two sorted arrays. The overall run time compl…

Median of Two Sorted Arrays 搜索时间复杂度的时候,看到归并排序比较适合这个题目.中位数直接取即可,所以重点是排序. 再来看看治阶段,我们需要将两个已经有序的子序列合并成一个有序序列 by 图解排序算法(四)之归并排序 第1次提交 class Solution: def findMedianSortedArrays(self, nums1, nums2): """ :type nums1: List[int] :type nums2: List[in…

4. Median of Two Sorted Arrays 给定两个有序的整数序列.求中位数,要求复杂度为对数级别. 通常的思路,我们二分搜索中位数,对某个序列里的某个数 我们可以在对数时间内通过二分算法求得两个序列中比它小的数,整体复杂度也是对数级别.但是代码实现较为困难. 换一个思路,我们把中位数不要当作一个数,而当作一个序列的划分.划分后,序列的左半部设为L,右半部设为R 满足max(L)<=min(R)且满足len(L)==len(R) 二分搜索这个划分即可.对于A+B的长度为奇数的情…

4. Median of Two Sorted Arrays Hard There are two sorted arrays nums1 and nums2 of size m and n respectively. Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)). You may assume nums1 and nums2 cannot be b…

题目描述:Median of Two Sorted Arrays There are two sorted arrays A and B of size m and n respectively. Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)). 分析: 题目中的数组A和数组B都是排好序的,首先想到的算法是:设置一个计数器m,指针pA指向数组A的首地址…

https://leetcode.com/problems/median-of-two-sorted-arrays/ There are two sorted arrays nums1 and nums2 of size m and n respectively. Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)). Example 1: nums1 =…

这是我做的第二个leetcode题目,一开始以为和第一个一样很简单,但是做的过程中才发现这个题目非常难,给人一种“刚上战场就踩上地雷挂掉了”的感觉.后来搜了一下leetcode的难度分布表(leetcode难度及面试频率)才发现,该问题是难度为5的问题,真是小看了它!网上搜了很多答案,但是鲜见简明正确的解答,唯有一种寻找第k小值的方法非常好,在此整理一下. 首先对leetcode的编译运行吐槽一下:貌似没有超时判断,而且small和large的数据集相差很小.此题一开始我采用最笨的方法去实现,利…

# -*- coding: utf8 -*-'''https://oj.leetcode.com/problems/median-of-two-sorted-arrays/ There are two sorted arrays A and B of size m and n respectively. Find the median of the two sorted arrays.The overall run time complexity should be O(log (m+n)).…

There are two sorted arrays nums1 and nums2 of size m and n respectively. Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)). Example 1: nums1 = [1, 3] nums2 = [2] The median is 2.0 Example 2: nums1 = [1,…

There are two sorted arrays nums1 and nums2 of size m and n respectively. Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)). You may assume nums1 and nums2 cannot be both empty. Example 1: nums1 = [1, 3]…

题目简述: There are two sorted arrays A and B of size m and n respectively. Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)). 解题思路: 这本身是个很简单的题目,但是题目要求他的复杂度为O(log(m+n)),就很有难度了.不过首先我们还是可以明确我们要用分治法.关键是怎么分治呢?我们…

题目描述: There are two sorted arrays nums1 and nums2 of size m and n respectively. Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)). 解题思路: 本题要求求解的是两个有序序列的中位数.本质上就是求两个有序序列“第k小的数“的变形.假设两个有序序列一个长为m,另一个长为n,则我们…

There are two sorted arrays nums1 and nums2 of size m and n respectively. Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)). 解题思路: 由于要求时间复杂度O(log (m+n))所以几乎可以肯定是递归和分治的思想. <算法导论>里有找两个数组第K小数的算法,时间复杂度为O(lo…

There are two sorted arrays A and B of size m and n respectively. Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)). 思路: 难,知道用分治算法,却不知道怎么用.只好看答案. 基本的思路是如果中位数是第K个数,A[i]如果是中位数,那么A[i]已经大于了i个数,还应大于K - i - 1个…

题目: There are two sorted arrays nums1 and nums2 of size m and n respectively. Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)). 题意: 两个排序后的数组nums1 和nums2,长度分别是m,n,找出其中位数,并且时间复杂度:O(log(m+n)) 最愚蠢的方法: 两个数组合…

问题描述: There are two sorted arrays nums1 and nums2 of size m and n respectively. Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)). 解题思路: 看到时间复杂度的时候就知道这种应该使用二分查找法了,否则如果实现log的时间复杂度? 思路已经有大神提供了,说的非常清楚,附上链接地…

题目:there are two sorted arrays nums1 and nums2 of size m and n respectively. Find the median of the two sorted arrays. The overall run time complexity should be O(log(m+n)) 题解: 1.自己想得思路:构建一个list,然后比较各数的大小,将其插入到合适的位置 class Solution: # @param {integer[…

题目: There are two sorted arrays nums1 and nums2 of size m and n respectively. Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)). Example 1: nums1 = [1, 3] nums2 = [2] The median is 2.0 Example 2: nums1 =…

There are two sorted arrays nums1 and nums2 of size m and n respectively. Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)). You may assume nums1 and nums2 cannot be both empty. Example 1: nums1 = [1, 3]…

[题目描述] There are two sorted arrays nums1 and nums2 of size m and n respectively. Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)). Example 1: nums1 = [1, 3] nums2 = [2] The median is 2.0 Example 2: nums…

传送门 Description There are two sorted arrays nums1 and nums2 of size m and n respectively. Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)). Example 1: nums1 = [1, 3] nums2 = [2] The median is 2.0 Exampl…

一.题目链接:https://leetcode.com/problems/median-of-two-sorted-arrays 二.题目大意: 给定两个排序过的数组,求出两个数组的中位数,要求时间复杂度为O(log(m+n)). 三.题解: 这是我在LeetCode上做的第一道hard级别的题目,没想到难度系数达到了5!被虐哭了...言归正传,这道题目的解题思路看起来并不难,难点在于它限定了时间复杂度,而且是对数级别的复杂度(说明优化程度已经相当高了).这里,我给出三种解决方案,当然前两种时间…

一.题目大意 标签: 查找 https://leetcode.cn/problems/median-of-two-sorted-arrays 给定两个大小分别为 m 和 n 的正序(从小到大)数组 nums1 和 nums2.请你找出并返回这两个正序数组的 中位数 . 算法的时间复杂度应该为 O(log (m+n)) . 示例 1: 输入:nums1 = [1,3], nums2 = [2] 输出:2.00000 解释:合并数组 = [1,2,3] ,中位数 2 示例 2: 输入:nums1 =…

思路:设现在可用区间在nums1是[s1,t1),nums2:[s2,t2) 1.当一个数组可用区间为0的时候,由于另一个数组是已经排过序的,所以直接可得 当要取的是最小值或最大值时,也直接可得 2.明显两个数组总长度为偶数的时候需要取最中间两个元素/2.0,长度为奇数时,只需要求最中间那个.所以只需要分别求出最多两个元素,分别是 (findKthElement(0,t1,0,t2,(t1 + t2)/2) 和 findKthElement(0,t1,0,t2,(t1 + t2)/2 + 1)…

貌似是去年阿里巴巴c++的笔试题,没有什么创新直接照搬的... 题意就是找出两个排序数组的中间数,其实就是找出两个排序数组的第k个数. 二分答案,先二分出一个数,再用二分算出这个数在两个排序数组排序第几,然后和k做比较,最后以这个比较为依据接着二分直到求出第k个数. 本来这是两重二分,由于楼主的懒已经没有办法治疗了,用库函数upper_bound()代替了第二重二分,有志者可以自己写第二重二分,楼主就偷懒了... 警告:由于题目很神奇,会出现两个排序数组为空的情况,楼主差点因为这个报警....…

题目: 找两个排序数组A[m]和B[n]的中位数,时间复杂度为O(log(m+n)). 解法: 更泛化的,可以找第k个数,然后返回k=(m+n)/2时的值. 代码: class Solution { public: double findMedianSortedArrays(int A[], int m, int B[], int n) { int total = m + n; if(total & 0x1) //总数为奇数,返回中位数 + ); else //总数为偶数,返回中间两个数的平局值…

两个有序的数组 nums1 和 nums2 维数分别为m,n.找所有数的中位数,复杂度 O(log (m+n)) 注意:奇偶个数,分治法求解,递归出口特殊处理.取Kth smallest数时,分治取mid=k/2和k-mid,避免奇数造成影响. class Solution { double findKth(vector<int> num1,vector<int> num2, int k) { int m = num1.size(), n = num2.size(); if(m &…