【重点】24.两两交换链表中的节点

题目

Python

参考：https://leetcode.cn/problems/swap-nodes-in-pairs/?envType=study-plan-v2&envId=top-100-liked![在这里插入图片描述](https://i-blog.csdnimg.cn/direct/8f3dc1face38497a9b580abd89304270.png

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 == None or head.next == None:
            return head
        newHead = head.next
        head.next = self.swapPairs(newHead.next)
        newHead.next = head
        return newHead

迭代

# 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 or not head.next:
            return head
        dummy = ListNode(-1)
        dummy.next = head
        pre = dummy
        cur = pre.next
        while cur and cur.next:
            nxt = cur.next.next
            pre.next = cur.next
            cur.next.next = cur
            cur.next = nxt
            pre = pre.next.next
            cur = pre.next
        
        return dummy.next

Java

法1：迭代

class Solution {
    public ListNode swapPairs(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }
        ListNode pre = null, cur = head, newHead = null;
        while (cur != null) {
            ListNode next = cur.next;
            if (next == null) { // 奇数个结点数量
                break;
            }
            ListNode tmp = next.next; // 下一组的起点
            if (pre == null) {
                next.next = cur;
                cur.next = tmp;
                newHead = next; // 新起点
            } else {
                pre.next = next;
                next.next = cur;
                cur.next = tmp;
            }
            pre = cur;
            cur = tmp;
        }

        return newHead;
    }
}

法2：迭代

跟迭代版翻转链表，异曲同工之妙！！！

class Solution {
    public ListNode swapPairs(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }
        ListNode newHead = head.next;
        head.next = swapPairs(newHead.next);
        newHead.next = head;
        return newHead;
    }
}