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.
一刷
题解：双指针first和second, 记住要保留first的prev指针信息
public class Solution {
public ListNode swapPairs(ListNode head) {
if(head == null || head.next == null) return head;
ListNode prev = head, first = head, second = head.next;
//ListNode third = null;
while(second!=null){
if(first == head) head = second;
else prev.next = second;
//third = second.next;
first.next = second.next;
second.next = first;
prev = first;
//
first = first.next;
second = first==null? null: first.next;
}
return head;
}
}
二刷：
recursive
/**
* 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 n = head.next;
head.next = swapPairs(head.next.next);
n.next = head;
return n;
}
}
