Swap Nodes in Pairs 解答

Question

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.

Solution

这题的核心在于dummy node和list操作。

dummy -> 1 -> 2 -> 3 -> 4

prev　　　cur   next  tmp

我们用四个指针来完成操作。

1. 预存tmp: tmp = next.next

2. 更改next: next.next = cur

3. 更改cur: cur.next = tmp

4. 更改prev: prev.next = next

5. 更新prev, cur, next:

cur = tmp

if (cur != null): next = cur.next

prev = prev.next.next

最后返回dummy.next

我们看到循环结束的条件应该是tmp为null或者tmp.next为null.

 /**
  * 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;
         }
         ListNode dummy = new ListNode(-1);
         dummy.next = head;
         ListNode prev = dummy, current = head, next = head.next, tmp;
         while (next != null) {
             tmp = next.next;
             next.next = current;
             current.next = tmp;
             prev.next = next;
             current = tmp;
             prev = prev.next.next;
             if (current == null) {
                 break;
             } else {
                 next = current.next;
             }
         }
         return dummy.next;
     }
 }