https://leetcode.com/problems/merge-two-binary-trees/ https://leetcode.com/problems/maximum-binary-tree/ https://leetcode.com/problems/maximum-width-of-binary-tree/ https://leetcode.com/problems/validate-binary-search-tree/ https://leetcode.com/problems/same-tree/ https://leetcode.com/prob...
题目地址:https://leetcode.com/problems/merge-sorted-array/description/ 题目描述 Given two sorted integer arrays nums1 and nums2, merge nums2 into nums1 as one sorted array. Note: The number of elements initialized in nums1 and nums2 are m and n respectively. You may assume that nums1 h...
题目地址:https://leetcode.com/problems/merge-sorted-array/description/ 题目描述 Given two sorted integer arrays nums1 and nums2, merge nums2 into nums1 as one sorted array. Note: The number of elements initialized in nums1 and nums2 are m and n respectively. You may assume...
链接:https://leetcode-cn.com/problems/merge-k-sorted-lists 终于来到进阶版本了,这道题说是hard,其实就是从归并排序(merge sort)演化来的,再+分治的思想 # Definition for singly-linked list. # class ListNode: # def __init__(self, x): # self.val = x # self.next = None classSolution: de...
// 参考:// 1)https://leetcode.cn/problems/merge-sorted-array/solution/he-bing-liang-ge-you-xu-shu-zu-by-leetco-rrb0/// 思路:// 1)状态初始化:nums1Index = m - 1,// nums2Index = n - 1, fillIndex = m + n - 1 。// 2)核心:循环处理,条件为 nums1Index >= 0 || nums2...
// 参考:// 1)https://leetcode.cn/problems/merge-sorted-array/solution/he-bing-liang-ge-you-xu-shu-zu-by-leetco-rrb0/// 思路:// 1)状态初始化:nums1Index = m - 1,// nums2Index = n - 1, fillIndex = m + n - 1 。// 2)核心:循环处理,条件为 nums1Index >= 0 || nums2...
来源:力扣(LeetCode) 链接:https://leetcode-cn.com/problems/merge-sorted-array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。 代码语言:javascript 复制 示例1: 输入:nums1=[1,2,3,0,0,0],m=3,nums2=[2,5,6],n=3输出:[1,2,2,3,5,6]解释:需要合并[1,2,3]和[2,5...
voidmerge(vector<int>&nums1,intm,vector<int>&nums2,intn){vector<int>res;inta=0;intb=0;for(inti=0;i<m+n;i++){if(a<m){if(b<n){if(nums1[a]<nums2[b]){res.push_back(nums1[a]);a++;}else{res.push_back(nums2[b]);b++;}}else{res.push_back(nums1[a]);a++;}}else...
721. 账户合并leetcode-cn.com/problems/accounts-merge/ 题目还是很好理解的,难就难在参透题目的本质,本题实质上是要判断,所有的邮箱地址中,哪些是属于同一个人的,然后将这个人找出来。 这样我们大体上可以分为两步: 第一步,找出哪些邮箱是一个分组的 ...
#链接:https://leetcode-cn.com/problems/merge-k-sorted-lists/solution/xiong-mao-shua-ti-python3-3chong-jie-fa-bao-li-you/ 这里我直接贴人家的代码来做对比,由于合并两个链表的函数结果是一个链表,那么就可以使用上面这种 合并(合并(一半),合并(另一半)) 的巧妙思路,调用次数瞬间降到了 log2(n)。