Description
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
Iteration
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode swapPairs(ListNode head) {
ListNode dummy = new ListNode(0);
dummy.next = head;
ListNode prev = dummy;
ListNode curr = head;
while (curr != null && curr.next != null) {
ListNode post = curr.next;
prev.next = post;
curr.next = post.next;
post.next = curr;
prev = curr;
curr = curr.next;
}
return dummy.next;
}
}
Recursion
其实使用两个指针就够了呀。
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode swapPairs(ListNode head) {
if (head == null || head.next == null) return head;
ListNode curr = head.next;
head.next = swapPairs(curr.next);
curr.next = head;
return curr;
}
}
