Given a linked list, swap every two adjacent nodes and return its head.
You may not modify the values in the list's nodes, only nodes itself may be changed.
Example:
Given 1->2->3->4, you should return the list as 2->1->4->3.
题意：
给定一个链表，每两个节点做一下交换。
Solution1: Two Pointers(fast & slow)
(1) use Two Pointers(fast & slow) to find a pair
(2)swap
(3) move forward to find a new pair
code:
1 /* 2 Time: O(n) 3 Space: O(1) 4 */ 5 class Solution { 6 public ListNode swapPairs(ListNode head) { 7 ListNode dummy = new ListNode(-1); 8 dummy.next = head; 9 head = dummy; 10 while (head.next != null && head.next.next != null) { 11 // narrow to find a pair 12 ListNode slow = head.next; 13 ListNode fast = head.next.next; 14 15 // swap 16 head.next = fast; 17 slow.next = fast.next; 18 fast.next = slow; 19 20 // move to next pair 21 head = slow; 22 } 23 return dummy.next; 24 } 25 }
转载于:https://www.cnblogs.com/liuliu5151/p/10674640.html
