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
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:...
*/publicclassEasy_21_MergeTwoSortedList{publicstaticvoidmain(String[] args){Easy_21_MergeTwoSortedListinstance=newEasy_21_MergeTwoSortedList();ListNodel1=newListNode(1);ListNodel2=newListNode(2);ListNodel3=newListNode(4); l1.next = l2; l2.next = l3; System.out.println(instance.listNodeToStri...
需要遍历 list2 中的全部 O(|list2|) 个结点 空间复杂度:O(1) 只需要维护常数个额外变量 代码(Python3) # Definition for singly-linked list. # class ListNode: # def __init__(self, val=0, next=None): # self.val = val # self.next = next class Solution: def mergeTwoLists(self, list...
l2.Next = mergeTwoLists(l1, l2.Next);returnl2; } } 执行结果: leetcode-cn: 执行用时:0ms, 在所有 Go 提交中击败了100.00%的用户内存消耗:2.6MB, 在所有 Go 提交中击败了26.43%的用户leetcode: Runtime:0ms, faster than100.00% of Go online submissions for Merge Two Sorted Lists.Memory Usage...
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 1. 2. # Definition for singly-linked list. ...
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] Example 2: ...
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. 依次拼接 复杂度 时间O(N) 空间 O(1) 思路 该题就是简单的把两个链表的节点拼接起来,我们可以用一个Dummy头,将比较过后的节点接在这个Dummy头之后。
用一个大小为K的最小堆(用优先队列+自定义降序实现)(优先队列就是大顶堆,队头元素最大,自定义为降序后,就变成小顶堆,队头元素最小),先把K个链表的头结点放入堆中,每次取堆顶元素,然后将堆顶元素所在链表的下一个结点加入堆中。 代码语言:javascript ...
class ListNode: def __init__(self, x): self.val = x self.next = None class Solution: def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode: head = ListNode(0) first = head while l1 and l2: if l1.val > l2.val: head.next = l2 l2 = l2.next else: head.next =...