노트/Algorithm : 알고리즘
[leetcode] 24. Swap Nodes in Pairs
Diane_
2022. 11. 18. 10:06
Given a linked list, swap every two adjacent nodes and return its head. You must solve the problem without modifying the values in the list's nodes (i.e., only nodes themselves may be changed.)
Example 1:
Input: head = [1,2,3,4]
Output: [2,1,4,3]
Example 2:
Input: head = []
Output: []
Example 3:
Input: head = [1]
Output: [1]
Constraints:
- The number of nodes in the list is in the range [0, 100].
- 0 <= Node.val <= 100
# Definition for singly-linked list.
class ListNode(object):
def __init__(self, val=0, next=None):
self.val = val
self.next = next
class Solution(object):
def swapPairs(self, head):
"""
:type head: ListNode
:rtype: ListNode
"""
node = head
while node and node.next:
node.val, node.next.val= node.next.val, node.val
node = node.next.next
return head
class Solution(object):
def swapPairs(self, head):
"""
:type head: ListNode
:rtype: ListNode
"""
root = prev = ListNode(None)
prev.next = head
while head and head.next:
# b가 a(head)를 가리키도록 할당
b = head.next
head.next = b.next
b.next = head
# prev가 b를 가리키도록 할당
prev.next = b
# 다음번 비교를 위해 이동
head = head.next
prev = prev.next.next
return root.next