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 maynot modify the values in the list, only nodes itself can be changed.
解题思路：
就是不断地交换相邻两个节点的next，不断循环下去直到结束，记得就是在交换当前两个节点的next（例如k跟k+1）时，记得更新k-1的next。
[java] view plain copy print ? /** * 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; //当前节点k ListNode p = head; //下一节点k+1 ListNode q = head.next; //记录上一节点k-1 ListNode r = null; head = q; while(p != null && q != null) { //交换next p.next = q.next; q.next = p; //更新上一节点的next if(r != null) r.next = q; //更新节点 r = p; p = p.next; if(p != null) q = p.next; } return head; } } /** * 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; //当前节点k ListNode p = head; //下一节点k+1 ListNode q = head.next; //记录上一节点k-1 ListNode r = null; head = q; while(p != null && q != null) { //交换next p.next = q.next; q.next = p; //更新上一节点的next if(r != null) r.next = q; //更新节点 r = p; p = p.next; if(p != null) q = p.next; } return head; } }
