LeetCode-24. 两两交换链表中的节点【递归 链表】

题目描述：

给你一个链表，两两交换其中相邻的节点，并返回交换后链表的头节点。你必须在不修改节点内部的值的情况下完成本题（即，只能进行节点交换）。

示例 1：

输入：head = [1,2,3,4]
输出：[2,1,4,3]
示例 2：

输入：head = []
输出：[]
示例 3：

输入：head = [1]
输出：[1]

提示：

链表中节点的数目在范围 [0, 100] 内
0 <= Node.val <= 100

解题思路一：双指针1

# 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: Optional[ListNode]) -> Optional[ListNode]:
        if not head: return head
        dummy_head = ListNode(next=head)
        pre = dummy_head
        cur = dummy_head.next
        while cur.next:
            temp = cur.next
            pre.next = temp
            cur.next = temp.next
            temp.next = cur
            if cur.next == None:
                break
            pre = cur
            cur = cur.next

        return dummy_head.next

时间复杂度：O(n)
空间复杂度：O(1)

解题思路二：双指针2

# 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: Optional[ListNode]) -> Optional[ListNode]:
        if head == None or head.next == None: return head
        dummyHead = ListNode()
        dummyHead.next = head
        cur = dummyHead
        while cur.next != None and cur.next.next != None:
            tmp = cur.next
            tmp1 = cur.next.next.next

            cur.next = cur.next.next
            cur.next.next = tmp
            cur.next.next.next = tmp1

            cur = cur.next.next # 每次前进两个，并且判断后面有两个元素就交换
        return dummyHead.next

时间复杂度：O(n)
空间复杂度：O(1)

解题思路三：递归

# 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: Optional[ListNode]) -> Optional[ListNode]:
        if head is None or head.next is None: return head

        pre = head
        cur = head.next
        next = head.next.next

        cur.next = pre
        pre.next = self.swapPairs(next)
        return cur

时间复杂度：O(n)
空间复杂度：O(n)