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. Example: Input:1->2->4, 1->3->4Output:1->1->2->3->4->4 因为没有空间要求,所以想到ListNode*head = new ListNode(INT_MIN);重新定义一...
https://leetcode.com/problems/merge-two-sorted-lists/ 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. Example: Input: 1->2->4, 1->3->4 Output: 1->1->2->3->4->4 回到顶部 Intuitio...
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. 合并2个有序链表 2、代码实现 AI检测代码解析 /** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; ...
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. Seen this question in a real interview before? Yes 简单的归并排序,不说了,直接上代码:
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. Input: 1->3->5->null 2->4->6->7->null Output:1->2->3->4->5->6->7->null Day 1944 答案揭晓 DS Interview Question & Answer Expla...
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. Input: 1->3->5->null 2->4->6->7->null Output:1->2->3->4->5->6->7->null Day 1944 答案揭晓 DS Interview Question & Answer Expla...
next: return head # Nodes to be swapped first_node = head second_node = head.next # Swapping first_node.next = self.swapPairs(second_node.next) second_node.next = first_node # Now the head is the second node return second_node #作者:LeetCode #链接:https://leetcode-cn.com/problems...
方法1:使用优先队列合并 我们需要维护当前每个链表没有被合并的元素的最前面一个,k个链表就最多有 k个满足这样条件的元素,每次在这些元素里面选取 val 属性最小的元素合并到答案中。在选取最小元素的时候,我们可以用优先队列来优化这个过程。 代码语言:javascript ...
链接:https://leetcode-cn.com/problems/merge-two-sorted-lists/ Link:https://leetcode.com/problems/merge-two-sorted-lists/ 双指针 O(N) 比较两个链表头,把小的拿出来放到新的链表中 # Definition for singly-linked list.# class ListNode:# def __init__(self, val=0, next=None):# self.val...
next return head # 将链表转化为数组的函数,不需要改动 def ll_to_list(head): list = [] while head: list.append(head.val) head = head.next return list class Solution: def mergeTwoLists(self, head1, head2): if not head1: # 如果head1为空,直接返回head2 return head2 if not head2:...