[leetcode] Swap Nodes in Pairs

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.

https://oj.leetcode.com/problems/swap-nodes-in-pairs/

思路：模拟题，仔细操作即可。

1. 增加dummyhead方便处理；
2. 操作节点需保留前一个节点的指针。
   

代码：

public class Solution {
    public ListNode swapPairs(ListNode head) {
        if (head == null)
            return null;
        ListNode dummyHead = new ListNode(-1);
        dummyHead.next = head;
        ListNode pre = dummyHead;
        while (pre != null && pre.next != null && pre.next.next != null) {
            ListNode q = pre.next;
            ListNode r = q.next;
            pre.next = r;
            q.next = r.next;
            r.next = q;
            pre = pre.next.next;
        }
        return dummyHead.next;

    }

    public static void main(String[] args) {
        ListNode list = ListUtils.makeList(1, 2, 3, 4, 5);
        ListUtils.printList(list);
        ListUtils.printList(new Solution().swapPairs(list));
    }

}

第二遍记录：注意pre每次迭代是 pre=pre.next.next; 而不是pre=post，因为post已经变了，开始这里搞错了。

public class Solution {
    public ListNode swapPairs(ListNode head) {
        if(head==null)
            return null;
        ListNode dummyHead = new ListNode(-1);
        dummyHead.next=head;
        ListNode pre = dummyHead;
        while(pre!=null&&pre.next!=null&&pre.next.next!=null){
            ListNode cur = pre.next;
            ListNode post = cur.next;
            
            pre.next =post;
            cur.next=post.next;
            post.next=cur;
            
            pre=pre.next.next;
        }
        return dummyHead.next;   
    }
}