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...
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 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 class Solution: # @param {integer[]} num...
这道算法题属于LeetCode中hard级别的题目,难度主要在于确定二分查找对象和极端边界情况的处理。但是一旦将这两点考虑透彻,写代码将如砍瓜切菜一般。 最后祝各位同学面试超常发挥,毕业生找个好工作,想跳槽的人工资翻倍 !! xcode 算法细节系列(8):4. Median of Two Sorted Arrays ...
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…
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 [...
https://leetcode-cn.com/problems/median-of-two-sorted-arrays/solution/4-xun-zhao-liang-ge-you-xu-shu-zu-de-zhong-wei-shu/ class Solution { public double findMedianSortedArrays(int[] nums1, int[] nums2) { int n = nums1.length; ...
classSolution{ public: doublefindMedianSortedArrays(intA[],intm,intB[],intn) { inttotal=m+n; if(total&0x1) returnfind_kth(A,m,B,n,total/2+1); else return(find_kth(A,m,B,n,total/2) +find_kth(A,m,B,n,total/2+1))/2.0; ...
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]){...
陷阱:数组下标计算。另外由于LeetCode网站测试用例设置问题,线性时间复杂度的算法也可能被接受。 代码: classSolution:# @param {integer[]} nums1# @param {integer[]} nums2# @return {float}deffind(self,nums1,nums2,k):ifnotnums1:returnnums2[k]ifnotnums2:returnnums1[k]ia,ib=len(nums1)//2,...