【Leetcode】24. Swap Nodes in Pairs

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.

Tips：给定一个链表，在不改变链表的值，只改变结点的指针的情况下完成相邻两个结点的交换。

如：Input：

1->2->3->4

　　Output：

2->1->4->3

思路：设置两个指针，first second，分别为当前指针的next以及next.next。

将 first.next指向second.next;（1->3）

然后将second.next指向first(2->1)。

再将当前指针后移：cur.next=second; cur = first;

package medium;

import dataStructure.ListNode;

public class L24SwapNodesInPairs {

public ListNode swapPairs(ListNode head) {
if (head == null)
return head;
ListNode node = new ListNode(0);
node.next = head;
ListNode cur = node;

while (cur.next != null && cur.next.next!=null) {
ListNode first = cur.next;
ListNode second = cur.next.next;
first.next = second.next;//1->3
second.next=first;//2->1->3
cur.next=second;
cur = first;
}
return node.next;

}

public static void main(String[] args) {
ListNode head1 = new ListNode(1);
ListNode head2 = new ListNode(2);
ListNode head3 = new ListNode(3);
ListNode head4 = new ListNode(4);
ListNode head5 = new ListNode(5);
ListNode head6 = new ListNode(6);
ListNode head7 = null;
head1.next = head2;
head2.next = head3;
head3.next = head4;
head4.next = head5;
head5.next = head6;
head6.next = head7;
L24SwapNodesInPairs l24 = new L24SwapNodesInPairs();
ListNode head = l24.swapPairs(head1);
while (head != null ) {
System.out.println(head.val);
head=head.next;
}
}
}