next = list2; } // 返回合并后的链表的头结点 head_pre.next } } 题目链接: Merge Two Sorted Lists : leetcode.com/problems/m 合并两个有序链表: leetcode-cn.com/problem LeetCode 日更第 52 天,感谢阅读至此的你 欢迎点赞、收藏鼓励支持小满...
## 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...
*/functionmergeTwoLists($l1,$l2){if($l1==null) {return$l2; }if($l2==null) {return$l1; }if($l1->val <=$l2->val) {$l1->next =mergeTwoLists($l1->next,$l2);return$l1; }if($l2->val <=$l1->val) {$l2->next =mergeTwoLists($l2->next,$l1);return$l2; } }/** * 构...
list2->next = mergeTwoLists(list1, list2->next);returnlist2; } } }; Java 解法二: publicclassSolution {publicListNode mergeTwoLists(ListNode list1, ListNode list2) {if(list1 ==null)returnlist2;if(list2 ==null)returnlist1;if(list1.val <list2.val) { list1.next=mergeTwoLists(list1...
Can you solve this real interview question? Merge Two Sorted Lists - You are given the heads of two sorted linked lists list1 and list2. Merge the two lists into one sorted list. The list should be made by splicing together the nodes of the first two li
LeetCode-cn Merge two sorted linked lists and return it as a sorted list. The list should be made by splicing together the nodes of the first two lists. Example 1: Input: l1 = [1,2,4], l2 = [1,3,4] Output: [1,1,2,3,4,4] ...
https://leetcode.com/problems/merge-two-sorted-lists/ 题目: Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists. 思路: easy 。 算法: 1. public ListNode mergeTwoLists(ListNode l1, ListNode l2) {...
1. 题目描述 Merge two sorted linked lists and return it as a new sorted list. The new list should be made by splicing together the nodes of the first two lists. Example: Input:1->2->4,1->3->4Output:1->1->2->3->4->4 ...
[LeetCode OJ]- Merge Two Sorted Lists 题目要求:合并两个单向已排序的链表l1和l2,返回新的链表。 思路:该问题跟合并两个已排序的数组很像,合并两个已排序的数组是使用三个指针,从尾向头扫描,不断加入到数组中。而单向链表不能从尾往头加,这时候考虑也用两个“指针”从头往尾扫,加入到第三个新的链表中...
vector<int> mergeKSortedArrays(vector<vector<int>> &kArrays, int k) { // making a min heap priority_queue<Node *, vector<Node *>, compare> minheap; // insert all first indexes of arrays into the heap for (int i = 0; i < k; i++) { Node *tmp = new Node(kArrays[i][0]...