Since A has m elements, so there are m+1 kinds of cutting( i = 0 ~ m ). And we know: len(left_A) = i, len(right_A) = m - i . Note: when i = 0 , left_A is empty, and when i = m , right_A is empty. With the same way, cut B into two parts at a random po...
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...
代码二 class Solution { public: double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) { if(nums1.size() > nums2.size()) return findMedianSortedArrays(nums2, nums1); if(nums1.size() == 0) { return (nums2[(nums2.size() - 1) / 2 ] + nums2[nums2.size()/...
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...
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:...
[LeetCode] 28. 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-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...
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
classSolution{public:intfindKth(int*a,intm,int*b,intn,intk){if(m>n)returnfindKth(b,n,a,m,k);if(m==0)returnb[k-1];if(k==1)returnmin(a[0],b[0]);inti=min(m,k/2),j=k-i;if(a[i-1]<b[j-1]){returnfindKth(a+i,m-i,b,n,k-i);}elseif(a[i-1]>b[j-1]){...
classSolution{publicdoublefindMedianSortedArrays(int[]nums1,int[]nums2){//构建新数组int[]nums=newint[nums1.length+nums2.length];//将数组合并(采用arraycopy函数)if(nums1.length>0&&nums2.length>0){System.arraycopy(nums1,0,nums,0,nums1.length);System.arraycopy(nums2,0,nums,nums1.length,...