[LEET CODE] 24. Swap Nodes in Pairs

# Problem

Given a linked list, swap every two adjacent nodes and return its head.   Constraints: The number of nodes in the list is in the range [0, 100]. 0 <= Node.val <= 100

 

# My Answer

# 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:
        if head is None or head.next is None : 
            return head
            
        # 처음 페어 바꾸기
        node = head.next
        node.next,node.next.next  = head,head.next.next
        head = node
        node = node.next
        
        while node.next and node.next.next:
            temp = node.next
            node.next = node.next.next
            node = node.next
            temp.next = node.next
            node.next = temp
            node = node.next
        
        return head

 

# Solution 1 - 값만 교환

# Definition for singly-linked list.
class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None


class Solution:
    def swapPairs(self, head: ListNode) -> ListNode:
        cur = head

        while cur and cur.next:
            # 값만 교환
            cur.val, cur.next.val = cur.next.val, cur.val
            cur = cur.next.next

        return head

 

# Solution 2 - 반복 구조로 스왑

  
# Definition for singly-linked list.
class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None


class Solution:
    def swapPairs(self, head: ListNode) -> ListNode:
        root = prev = ListNode(None)
        prev.next = head
        while head and head.next:
            # b가 a(head)를 가리키도록
            b = head.next
            head.next = b.next
            b.next = head

            # prev가 b를 가리키도록
            prev.next = b

            # 다음 번 비교를 위해 이동
            head = head.next
            prev = prev.next.next
        return root.next

 

# Solution 3 - 재귀 구조로 스왑

# Definition for singly-linked list.
class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None


class Solution:
    def swapPairs(self, head: ListNode) -> ListNode:
        if head and head.next:
            p = head.next
            # 스왑된 값 리턴 받음
            head.next = self.swapPairs(p.next)
            p.next = head
            return p
        return head