首页 > 编程语言 >24. Swap Nodes in Pairs

24. Swap Nodes in Pairs

时间:2022-11-04 09:56:21浏览次数:52  
标签:24 pre Pairs ListNode cur next dmy suc Nodes

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 1:

class Solution {

    public ListNode swapPairs(ListNode head) {

       ListNode dmy = new ListNode(0), pre = dmy;

       dmy.next = head;

       while (pre.next != null && pre.next.next != null) {

          ListNode cur = pre.next;

          ListNode suc = cur.next; 

          ListNode tmp = suc.next;

          pre.next = suc;

          suc.next = cur;

          cur.next = tmp;

          pre = cur;

       }

       return dmy.next;

   }

}

标签:24,pre,Pairs,ListNode,cur,next,dmy,suc,Nodes
From: https://www.cnblogs.com/MarkLeeBYR/p/16856718.html

相关文章