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题目来源

24. 两两交换链表中的节点 - 力扣（LeetCode）

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

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

题目解析

本题主要考察数据结构。

对于输入的链表，我们可以为其定义一个虚拟头节点 dummy_head，比如示例1，进行如下逻辑

C源码实现

/*** Definition for singly-linked list.* struct ListNode {*     int val;*     struct ListNode *next;* };*/
struct ListNode* swapPairs(struct ListNode* head) {struct ListNode* dummy_head = (struct ListNode*)malloc(sizeof(struct ListNode));dummy_head->val = 0;dummy_head->next = head;struct ListNode* pre = dummy_head;struct ListNode* cur = pre->next;while (cur != NULL && cur->next != NULL) { // 由于要交换cur和cur.next两个节点，因此二者不能为nullstruct ListNode* nxt = cur->next;cur->next = nxt->next;nxt->next = cur;pre->next = nxt;pre = cur;cur = pre->next;}return dummy_head->next;
}

C++源码实现

/*** Definition for singly-linked list.* struct ListNode {*     int val;*     ListNode *next;*     ListNode() : val(0), next(nullptr) {}*     ListNode(int x) : val(x), next(nullptr) {}*     ListNode(int x, ListNode *next) : val(x), next(next) {}* };*/
class Solution {
public:ListNode* swapPairs(ListNode* head) {ListNode* dummy_head = new ListNode(0, head);ListNode* pre = dummy_head;ListNode* cur = pre->next;while (cur != nullptr && cur->next != nullptr) { // 由于要交换cur和cur.next两个节点，因此二者不能为nullListNode* nxt = cur->next;cur->next = nxt->next;nxt->next = cur;pre->next = nxt;pre = cur;cur = pre->next;}return dummy_head->next;}
};

Java源码实现

/*** Definition for singly-linked list.* public class ListNode {*     int val;*     ListNode next;*     ListNode() {}*     ListNode(int val) { this.val = val; }*     ListNode(int val, ListNode next) { this.val = val; this.next = next; }* }*/class Solution {public ListNode swapPairs(ListNode head) {ListNode dummy_head = new ListNode(0, head);ListNode pre = dummy_head;ListNode cur = pre.next;while (cur != null && cur.next != null) { // 由于要交换cur和cur.next两个节点，因此二者不能为nullListNode nxt = cur.next;cur.next = nxt.next;nxt.next = cur;pre.next = nxt;pre = cur;cur = pre.next;}return dummy_head.next;}
}

Python源码实现

# 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: Optional[ListNode]:rtype: Optional[ListNode]"""dummy_head = ListNode(0, head)pre = dummy_headcur = pre.nextwhile cur and cur.next:  # 由于要交换cur和cur.next两个节点，因此二者不能为nullnxt = cur.nextcur.next = nxt.nextnxt.next = curpre.next = nxtpre = curcur = pre.nextreturn dummy_head.next

JavaScript源码实现

/*** Definition for singly-linked list.* function ListNode(val, next) {*     this.val = (val===undefined ? 0 : val)*     this.next = (next===undefined ? null : next)* }*/
/*** @param {ListNode} head* @return {ListNode}*/
var swapPairs = function (head) {const dummy_head = new ListNode(0, head);let pre = dummy_head;let cur = pre.next;while (cur != null && cur.next != null) { // 由于要交换cur和cur.next两个节点，因此二者不能为nullconst nxt = cur.next;cur.next = nxt.next;nxt.next = cur;pre.next = nxt;pre = cur;cur = pre.next;}return dummy_head.next;
};