[leetcode]Swap Nodes in Pairs

Swap Nodes in Pairs

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.

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 }

 

第二遍记录：

public class Solution {
    public ListNode swapPairs(ListNode head) {
        if(head == null || head.next == null) return head;
        ListNode hhead = new ListNode(0);
        hhead.next = head;
        ListNode pre = hhead;
        while(pre.next != null && pre.next.next != null){
            ListNode p = pre.next.next;
            pre.next.next = p.next;
            p.next = pre.next;
            pre.next = p;
            pre = pre.next.next;
        }
        return hhead.next;
    }
}

两个指针就够用了，上一遍太小心了