LeetCode2.2.8(Swap Nodes in Pairs)

原创于 2015-08-29 10:10:14 发布 · 277 阅读

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本文介绍了一种链表操作技巧，通过遍历链表并交换每对相邻节点的位置来解决问题，仅使用常量空间，不修改节点值。

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.

public static void solution_2_2_8(Node head){
		Node pre=null,p1=head,temp=null,p2=null,h=null;
		while(p1!=null){
			if(p1.next!=null){
				p2=p1.next;
				temp=p2.next;
				if(pre==null){
					h=p2;
					p1.next=p2.next;
					p2.next=p1;
					pre=p1;
				}
				else{
					pre.next=p2;
					p1.next=p2.next;
					p2.next=p1;
					pre=p1;
				}
				p1=temp;
			}
			else{
				p1=p1.next;
			}
		}
		
		for(Node r=h;r!=null;r=r.next){
			System.out.print(r.data+" ");
		}
	}