Leet Reorder List

最新推荐文章于 2024-11-20 21:28:18 发布

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本文介绍了一个LeetCode上的题目——重构链表的解决方案。该算法通过将链表分成两部分，再将第二部分反转并与第一部分交错连接来完成题目要求。详细介绍了实现步骤及Java代码细节。

https://oj.leetcode.com/problems/reorder-list/

这个题一开始我以为在recorderList函数里面输出record之后的序列，实际上这个函数只是要求把数据进行record就行了，head可以在这个函数外调用，类似于c++中的引用和

指针。还是对java不算熟啊。。。。

	public void reorderList(ListNode head) {
		 
        ListNode head1 = head,head2,head3;
        int n=0;
        while(head1!=null){
        	head1 = head1.next;
        	n++;
        }
        
        if(n<=2){
        	return;
        }
        if(n==3){
        	head1 = head.next;
        	head2 = head1.next;
        	head.next = head2;
        	head2.next = head1;
        	head1.next = null;
        	return;
        }
        
        int m;
        if(n%2==1){
        	m = n/2;
        }
        else{
        	m = n/2-1;
        }
        head1=head;
    	for(int i=1;i<=m;i++)
    		head1= head1.next;
    	head2 = head1.next;
    	head3 = head2.next;
    	head1.next = null;
    	head2.next = null;
    	while(head3!=null){
    		head1 = head2;
    		head2 = head3;
    		head3 = head3.next;
    		head2.next = head1;
    	}
    	head2.next = head1;
    	head1 = head;
    	head3 = new ListNode(0);
    	head = head3;
    	while(head1!=null&&head2!=null){
    		head3.next = head1;
    		head1 = head1.next;
    		head3 =head3.next;
    		
    		head3.next = head2;
    		head2 = head2.next;
    		head3 = head3.next;
    	}
    	while(head1!=null){
    		head3.next = head1;
    		
    		head3 = head3.next;
    		head1 = head1.next;
    	}
    	
    	while(head2!=null){
    		head3.next = head2;
    		
    		head3 = head3.next;
    		head2 = head2.next;
    	}
    	head3 = head.next;
    	head.next =null;
    	head = head3;
    }