Leetcode||24.Swap Nodes in Pairs

24. Swap Nodes in Pairs

Total Accepted: 133469
Total Submissions: 361910
Difficulty: Easy
Contributors: Admin

Given a linked list, swap every two adjacent nodes and return its head.

For example,

Given

1->2->3->4

, you should return the list as

2->1->4->3

.

Your algorithm should use only constant space. You may not modify the values in the list, only nodes itself can be changed.

最近的链表真多，数据交换虽然不难，但是还是挺烦的。

代码如下：

/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode swapPairs(ListNode head) {
if (head == null || head.next == null) {
return head;
}
ListNode p,q,s,t;
q = head;
p = head.next;
if (p.next == null || p.next.next == null) {
head = p;
}
int count = 0;
while (p.next != null && p.next.next != null) {
s = p.next;
q.next = s;
p.next = q;
t = q;
if (count==0) {
head = p;
}
count++;
q = s;
p = q.next;
t.next = p;
}
t = p.next;
p.next = q;
q.next = t;
return head;
}
}