LeetCode刷题系列—4.Median of Two Sorted Arrays 1.题目描述 英文版: 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 ...
Code classSolution{public:doublefindMedianSortedArrays(vector<int>& a,vector<int>& b){// m,n表示两vector数组的大小。intm = a.size();intn = b.size();// 若a数组长度小于b数组,则交换两数组。因为我们要根据长数组元素定位短数组中的对应元素if(m < n)returnfindMedianSortedArrays(b,a);intl ...
Find the Index of the First Occurrence in a String 找出字符串中第一个匹配项的下 Grandyang刷尽天下 90 0 25:07 [LeetCode] 5. Longest Palindromic Substring 最长回文子串 Grandyang刷尽天下 113 0 13:44 [LeetCode] 15. 3Sum 三数之和 Grandyang刷尽天下 14 0 15:22 [LeetCode] 22. ...
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] ...
参考HuaHua酱的讲解:https://zxi.mytechroad.com/blog/algorithms/binary-search/leetcode-4-median-of-two-sorted-arrays/算法流程 假设序列nums1长度为n1 ,序列nums2长度n2 , 而且 n1 <= n2; 定位出中位数…
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...
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. 链接:https://leetcode-cn.com/problems/median-of-two-sorted-arrays 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
LeetCode Median of Two Sorted Arrays 找中位数(技巧),题意:给两个有序(升or降)的数组,求两个数组合并之后的中位数。思路:按照找第k大的思想,很巧妙。将问题的规模降低,对于每个子问题,k的规模至少减半。考虑其中一个子问题,在两个有序数组中找第k大,我们的
[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...
最后是Code publicdoublefindMedianSortedArrays(int[]A,int[]B){//获取长度intm=A.length;intn=B.length;//确保长度m<n(当然在这里我们也可以用递归来实现,不过效率会低一点)if(m>n){int[]temp=A;A=B;B=temp;inttmp=m;m=n;n=tmp;}//初始化变量intiMin=0...