[LeetCode] 24. 两两交换链表中的节点

题目链接:https://leetcode-cn.com/problems/swap-nodes-in-pairs/

题目描述:

给定一个链表，两两交换其中相邻的节点，并返回交换后的链表。

你不能只是单纯的改变节点内部的值，而是需要实际的进行节点交换。

示例:

给定 1->2->3->4, 你应该返回 2->1->4->3.

思路:

两种思路:迭代,递归

直接看代码理解!

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代码:

迭代

python(通过创建新的节点,不符合题意)

# 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:
        dummy = ListNode(0)
        p = dummy
        h = head
        while h:
            # 如果有接下来有两个节点
            if h and h.next:
                p.next = ListNode(h.next.val)
                p = p.next
                p.next = ListNode(h.val)
                h = h.next.next
            else:
                p.next = ListNode(h.val)
                h = h.next
            p = p.next
        return dummy.next

不创建新节点

python

# 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:
        dummy = ListNode(0)
        p = dummy
        h = head
        while h:
            if h and h.next:
                tmp = h.next
                p.next = tmp
                # 交换 位置
                h.next = h.next.next
                tmp.next = h
                h = h.next
                p = p.next.next
            else:
                p.next = h
                h = h.next
        return dummy.next

java

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode swapPairs(ListNode head) {
        ListNode dummy = new ListNode(0);
        ListNode p = dummy;
        while (head != null){
            if (head != null && head.next != null){
                ListNode tmp = head.next;
                p.next = tmp;
                head.next = head.next.next;
                tmp.next = head;
                head = head.next;
                p = p.next.next;
            }
            else {
                p.next = head;
                head = head.next;
            }
        }
        return dummy.next;
    }
}

递归

python

# 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 not head or not head.next:
            return head
        tmp = head.next
        head.next = self.swapPairs(head.next.next)
        tmp.next = head
        return tmp

java

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode swapPairs(ListNode head) {
        if (head == null || head.next == null) return head;
        ListNode tmp = head.next;
        head.next = swapPairs(head.next.next);
        tmp.next = head;
        return tmp;

    }
}