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)). 解法一:保底做法,O(m+n)复杂度 按照归并排序的思路,先归并,再找中间值。 classSolution {public:doublefindMedianSortedArrays...
classSolution {publicdoublefindMedianSortedArrays(int[] nums1,int[] nums2) {intm=nums1.length;intn=nums2.length;int[] A=nums1;int[] B=nums2;if(m>n){inttemp1=m; m=n; n=temp1;int[] temp2=A; A=B; B=temp2; }intmax_left=0;intmin_right=0;//i+j=m-i+n-j+1intimin=0,...
There are two sorted arraysnums1andnums2of size m and n respectively. Find the median of the two sortedarrays. 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, 2] nums2 = [3, 4] ...
[LeetCode] 1. Two Sum 两数之和 Grandyang刷尽天下 125 0 13:05 [LeetCode] 2. Add Two Numbers 两个数字相加 Grandyang刷尽天下 94 0 08:40 [LeetCode] 21. Merge Two Sorted Lists 合并两个有序链表 Grandyang刷尽天下 96 0 13:09 [LeetCode] 29. Divide Two Integers 两数相除 Grandy...
Code classSolution{public:doublefindMedianSortedArrays(vector<int>&nums1,vector<int>&nums2){intm=nums1.size(),n=nums2.size();if(m>n)returnfindMedianSortedArrays(nums2,nums1);intlow=0,high=m;while(left<=right){inti=low+(high-low)/2;intj=(m+n)/2-i;intleft1=(i==0)?INT_MIN:...
最后决定讲一道我个人觉得很有意思的一道题,那就是LeetCode上大名鼎鼎的Median of Two Sorted Arrays。这道题是一道非常经典并且有相当难度的二分查找问题,彻底理解和实现之后其他二分查找问题相信都会迎刃而解。 题目详情 原题如下: There are two sorted arrays nums1 and nums2 of size m and n respectively...
Leetcode-Hard 4. Median of Two Sorted Arrays 题目描述 有两个排序的数组nums1和nums2分别为m和n。 找到两个排序数组的中位数。总运行时间复杂度应为O(log(m + n))。 假设nums1和nums2不能都为空。 思路 将两个数组合并然后排序,根据合并后新数组长度来计算中位数...
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)). 题意已只两个有序的序列,找到他们的中位数,复杂度要求O(log (m+n))。
Median of Two Sorted Arrays问题:求2个有序数组的中位数,要求算法时间复杂度为O(log(m+n))。 难度:困难 思路:此题为“在2个有序数组中寻找第k个元素”问题的变体。根据中位数的定义,此题可转化为“在2个有序数组中寻找第‘(m+n)/2’个元素”。算法难点在于要求时间复杂度为O(log(m+n)),因此需要...
[i-1];}}doublefindMedianSortedArrays(vector<int>&nums1,vector<int>&nums2){intm=nums1.size(),n=nums2.size();if((m+n)%2){returnfindKth(nums1.data(),m,nums2.data(),n,(m+n)/2+1);}else{intleft=findKth(nums1.data(),m,nums2.data(),n,(m+n)/2);intright=findKth(nums1...