LeetCode24. 两两交换链表中的节点

题目

分析

利用哑结点方便操作。

head->n1->n2->n3->n4

交换步骤

1. n1指向n3

2. n2指向n1

3. head->next 指向n2

4. head指向n1（n3的前一个结点）

代码

 1 class Solution {
 2 public:
 3     ListNode* swapPairs(ListNode* head) {
 4         if(!head || !head->next) return head; //只有一个结点或者空结点
 5         ListNode *dummyHead = new ListNode(0);  //创造哑节点
 6         dummyHead->next = head;
 7         head = dummyHead;
 8         
 9         while(head && head->next && head->next->next){
10             auto n1 = head->next;
11             auto n2 = head->next->next;
12             //head->n1->n2-> ... ->
13             n1->next = n2->next;  //先操作n1防止断链
14             n2->next = n1;
15             head->next = n2;
16             head = n1;
17         }
18         return dummyHead->next;
19     }
20 };

 

 1 /**
 2  * Definition for singly-linked list.
 3  * struct ListNode {
 4  *     int val;
 5  *     ListNode *next;
 6  *     ListNode() : val(0), next(nullptr) {}
 7  *     ListNode(int x) : val(x), next(nullptr) {}
 8  *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 9  * };
10  */
11 class Solution {
12 public:
13     ListNode* swapPairs(ListNode* head) {
14         auto dummy = new ListNode(-1),cur = dummy;
15         dummy->next = head;
16         while(cur->next  && cur->next->next){
17             
18             auto p1 = cur->next,p2 = cur->next->next;
19             
20             cur->next = p2;
21             p1->next = p2->next;
22             p2->next = p1;
23 
24             cur = p1;
25         }
26         return dummy->next;
27     }
28 };

 

posted @ 2021-03-20 20:25  Uitachi  阅读(35)  评论(0)    收藏  举报