1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
|
/*
* @lc app=leetcode id=24 lang=java
*
* [24] Swap Nodes in Pairs
*
* https://leetcode.com/problems/swap-nodes-in-pairs/description/
*
* algorithms
* Medium (64.32%)
* Likes: 11816
* Dislikes: 436
* Total Accepted: 1.4M
* Total Submissions: 2.1M
* Testcase Example: '[1,2,3,4]'
*
* Given a linked list, swap every two adjacent nodes and return its head. You
* must solve the problem without modifying the values in the list's nodes
* (i.e., only nodes themselves may be changed.)
*
*
* Example 1:
*
*
* Input: head = [1,2,3,4]
* Output: [2,1,4,3]
*
*
* Example 2:
*
*
* Input: head = []
* Output: []
*
*
* Example 3:
*
*
* Input: head = [1]
* Output: [1]
*
*
*
* Constraints:
*
*
* The number of nodes in the list is in the range [0, 100].
* 0 <= Node.val <= 100
*
*
*/
// @lc code=start
/**
* 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; }
* }
思路：正常模拟就行，注意断开指针的顺序
*/
class Solution {
public ListNode swapPairs(ListNode head) {
//设置一个虚拟头节点
ListNode dumyHead = new ListNode(0);
//将虚拟头结点指向head，这样方便后面操作
dumyHead.next = head;
ListNode cur = dumyHead;
//临时节点，保存两个节点之中的第一个节点
ListNode firstNode ;
// 临时节点，保存两个节点之中的第二个节点
ListNode secondeNode ;
while (cur.next != null && cur.next.next != null) {
// tmpNode = cur.next.next.next;
firstNode = cur.next;
secondeNode = cur.next.next;
//1 cur指针next指向第2个节点
cur.next = secondeNode;
//2 第一个节点指针next指向第三个节点
firstNode.next = secondeNode.next;
//3 第二个节点指针next指向第一个节点
secondeNode.next = firstNode;
//移动cur指针，准备下一轮交换
cur = cur.next.next;
}
return dumyHead.next;
}
}
// @lc code=end
|
