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.
把一个链表中的每一对节点对换(不能只换值)
/** * 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 h = new ListNode(0); h.next = head; ListNode p = h; while(p.next != null && p.next.next != null){ //use t1 to track first node ListNode t1 = p; p = p.next; t1.next = p.next; //use t2 to track next node of the pair ListNode t2 = p.next.next; p.next.next = p; p.next = t2; } return h.next; } }
解题思路:这题主要涉及到链表的操作,没什么特别的技巧,注意不要出错就好。最好加一个头结点,操作起来会很方便。
# Definition for singly-linked list. # class ListNode: # def __init__(self, x): # self.val = x # self.next = None class Solution: # @param a ListNode # @return a ListNode def swapPairs(self, head): if head == None or head.next == None: return head dummy = ListNode(0); dummy.next = head p = dummy while p.next and p.next.next: tmp = p.next.next p.next.next = tmp.next tmp.next = p.next p.next = tmp p = p.next.next return dummy.next