标签：des style blog http color strong
Given a linked list, swap every two adjacent nodes and return its head.
For example,
Given1->2->3->4, you should return the list as2->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.
题中要求O(1)空间，且不允许修改list中节点的值。
事实上我尝试修改节点的值，还是过了，哈哈
算法思想：
每次选两个节点，同时需要维护第三个指针，就是这两个节点的前驱，以防链表断链。两节点互换很容易，换完之后要与前驱节点链接。
代码如下：
1 public class Solution { 2 public ListNode swapPairs(ListNode head) { 3 if(head == null || head.next == null) return head; 4 ListNode hhead = new ListNode(0); 5 hhead.next = head; 6 ListNode one = head; 7 ListNode two = one.next; 8 ListNode pre = hhead; 9 while(one !=null && two != null){ 10 one.next = two.next; 11 two.next = one; 12 pre.next = two; 13 pre = one; 14 one = one.next; 15 if(one == null) break; 16 two = one.next; 17 } 18 return hhead.next; 19 } 20 }
[leetcode]Swap Nodes in Pairs,布布扣,bubuko.com
标签：des style blog http color strong
原文地址：http://www.cnblogs.com/huntfor/p/3855170.html
