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.
这道题自己想的方法没有写出来，然后看了别人的解法。发现意思跟我的思路差不多，但细节很多地方我确实想不出来。linkedlist的题目很多细节问题，哪个该连哪个，对我来说现在很容易错或者陷入混乱。
该解法实际上仍然是维护三个pointers：prev, curt, temp.
以节点为偶数个时举例：
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当节点数为偶数的时候，是curt == null退出循环；当节点数为奇数的时候，是curt.next == null退出循环。
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/**
* 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 dummy = new ListNode(0);
dummy.next = head;
ListNode prev = dummy;
ListNode curt = head;
while (curt != null && curt.next != null){
ListNode temp = curt.next.next;
curt.next.next = curt;
prev.next = curt.next;
curt.next = temp;
prev = curt;
curt = curt.next;
}
return dummy.next;
}
}
