## 解法二:递归 Recursion class Solution: def mergeTwoLists(self, head1, head2): ## head1 和 head2 头结点指代两个链表 ## 递归的 base,结束条件 if head1 is None: return head2 elif head2 is None: return head1 elif head1.val < head2.val: ## 把小的值 head1 提取出来,拼接上后面...
LeetCode 88. Merge Sorted Array 程序员木子 香港浸会大学 数据分析与人工智能硕士在读 来自专栏 · LeetCode Description 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 ...
先复习下原本我们在MergeSort里面怎么利用一个新建的数量来merge two array: 代码如下: 1publicint[] mergeTwoList(int[] A,int[] B) { 2int[] C =newint[A.length + B.length]; 3intk = 0; 4inti = 0; 5intj = 0; 6while(i < A.length && j < B.length) { 7if(A[i] < B[j]) ...
力扣Leetcode 21|合并两个有序链表Merge Two sorted Array 2020年10月13日 12:00595浏览·3喜欢·0评论 爱学习的饲养员 粉丝:6.7万文章:46 关注 视频讲解 622:17 Leetcode力扣 1-300题视频讲解合集|手画图解版+代码【持续更新ing】 77.6万783 视频爱学习的饲养员 ...
【LeetCode】88. Merge Sorted Array (2 solutions) Merge Sorted Array Given two sorted integer arrays A and B, merge B into A as one sorted array. Note:You may assume that A has enough space (size that is greater or equal tom+n) to hold additional elements from B. The number of ...
按照归并排序的惯性思路,因为归并排序中给定的是一个数组的两个区间,所以通常情况下会借助O(n)大小的辅助空间。思路如下: class Solution { public: void merge(int A[], int m, int B[], int n) { int temp[m+n]; int i = 0, j = 0, t = 0; ...
Leetcode每日一题:88.merge-sorted-array(合并两个有序数组),思路:最容易想到的就是归并排序的归并策略了,时间复杂度O(m+n),空间复杂度O(n);但是官网给出了另一个难以想到的妙招,就是反向归并,因为nums1最后面的n个元素是空的,所以从后端开始归并,每次选择较大的放
Given two sorted integer arraysnums1andnums2, mergenums2intonums1as one sorted array. Note:You may assume thatnums1has enough space (size that is greater or equal tom+n) to hold additional elements fromnums2. The number of elements initialized innums1andnums2aremandnrespectively. ...
2021-01-21https://leetcode.com/problems/merge-sorted-array/ Description Given two sorted integer arrays nums1 and nums2, merge nums2 into nums1 as one sorted array. The number of elements initialized in nums1 and nums2 are m and n respectively. You may assume that nums1 has a size eq...
l1.next = self.mergeTwoLists(l1.next, l2) return l1 else: l2.next = self.mergeTwoLists(l1, l2.next) return l2 Test the the solution: s = Solution()l1 = ListNode(1, ListNode(4, ListNode(5)))l2 = ListNode(1, ListNode(3, ListNode(4)))l3 = ListNode(2, ListNode(6))lists = ...