【Leetcode】Swap Nodes in Pairs in JAVA 难得的一次写对不带改的。。附赠测试程序like always

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.

给一个链表,两两交换。

我的思路是两对两对看,但是分为三种情况,这对和后面的情形是:1-2-3-4;1-2-3;1-2;

然后根据不同的情况写出他们的交换策略~代码里有注释,看不懂的可以看注释明白我的思路~

附赠main函数供各位测试!!!

public class SwapNodesinPairs {
	public static void main(String args[]){
		SwapNodesinPairs dp = new SwapNodesinPairs();
		ListNode head  = new ListNode(1);
		ListNode p1 = new ListNode(2);
		head.next=p1;
		ListNode p2  = new ListNode(3);
		p1.next =  p2;
		ListNode p3 = new ListNode(4);
		p2.next = p3;
		ListNode p4 = new ListNode(5);
		p3.next = p4;
		prinf(head);
		prinf(dp.swapPairs(head));
	}

	private static void prinf(ListNode input){
		while(input!=null)	{
			System.out.print(input.val+"->");
			input = input.next;
		}
		System.out.println();
	}

	public ListNode swapPairs(ListNode head) {
        if(head==null||head.next==null)	return head;
        ListNode p=head;
        ListNode start = head.next;
        while(p!=null&&p.next!=null){
        	ListNode nxt = p.next;
        	//如果后面四个都有,2连向1,1连向4,4连向3,然后切记让下一轮的p得等于3(但是由于这个算法会跳过3,所以提前存好3)
        	if(p.next.next!=null&&p.next.next.next!=null){
        	p.next=nxt.next.next;
        	ListNode tmp = nxt.next;//提前存好3
        	nxt.next=p;
        	p=tmp;//让下一个p等于3,如果p.next的话下一个就是4了,而且3被永远跳过了
        	}
        	//如果后面有三个,就是这一对完了,还剩一个,那么最后那个不动,这两个交换
        	else if(p.next.next!=null&&p.next.next.next==null){
        		p.next=nxt.next;
        		nxt.next=p;
        		p=p.next;
        	}
        	//就剩最后两个,自己交换即可
        	else if(p.next.next==null){
        		nxt.next=p;
        		p.next=null;
        	}
        }
        return start;
    }
}
时间: 2024-11-10 14:37:31

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