Webthe first node in the pair. the second node in the pair. the node previous to the first node in the pair, so that its next field can be appropriately set.We make one pass through the list, modifying the pointers as desired: class Solution: def swapPairs (self, head: Optional [ListNode]) -> Optional [ListNode]: if not head: return None has_head ... Webdef swapPairs (self, head): dummy=ListNode(0) pre=dummy pre. next =head while pre. next and pre. next. next: a=pre. next b=a. next pre. next, a. next, b. next = b, b. next, a …

24 - Swap Nodes in Pairs Leetcode

WebFeb 7, 2024 · # Definition for singly-linked list. class ListNode: def __init__(self, val=0, next=None): self.val = val self.next = next class Solution: def swapPairs(self, head: ListNode) -> ListNode: # return original list if head is None or there's only one node in the list if head is None or head.next is None: return head # initialize current and previous … WebApr 10, 2024 · # Definition for singly-linked list. # class ListNode(object): # def __init__(self, val=0, next=None): # self.val = val # self.next = next class Solution … cle to sea alaska

leetcode/024_Swap_Nodes_in_Pairs.py at master - Github

Webdef insertAtHead(self, item): ''' pre: an item to be inserted into list post: item is inserted into beginning of list ''' node = ListNode(item) if not self.length: # set the cursor to the head … WebApr 9, 2024 · 解法二. 解决该问题需要两步：（1）判断链表是否有环（2）环的入口在哪里. 第一步先解决「是否有环」的问题：. 使用快慢指针 —— 快指针每次移动两步，慢指针 … WebJul 11, 2024 · class Solution: def swapPairs(self, head: ListNode) -> ListNode: prev = None curr = head while curr and curr.next: next = curr.next if prev is None: head = next … cle to spain