leetcode_19_Remove Nth Node From End of List

标签：链表   移动   倒数第n个结点   

描述：

Given a linked list, remove the nth node from the end of list and return its head.

For example,

   Given linked list: 1->2->3->4->5, and n = 2.

   After removing the second node from the end, the linked list becomes 1->2->3->5.

Note:
Given n will always be valid.
Try to do this in one pass.

思路：

先让p指向head，p往后移动n个节点，然后让q指向head，p和q一起后移直至p指向最后一个结点，则q指向的结点即是倒数第n个结点。当然，倒数第len（链表的长度）个结点是一个特殊情况，直接head=head.next即可。

代码：

public ListNode removeNthFromEnd(ListNode head, int n) {
        if(head==null)
        	return head;
        ListNode pListNode=head,qListNode=head;
        int len=0;
        while(pListNode!=null)
        {
        	len++;
        	pListNode=pListNode.next;
        }
        if(n==len)
        {
        	head=head.next;
        	return head;
        }
        pListNode=head;
        for(int i=0;i<n;i++)
        {
        	qListNode=qListNode.next;
        	if (qListNode==null)
        		break;
        }
        	
        while(qListNode!=null&&qListNode.next!=null)
        {
        	pListNode=pListNode.next;
        	qListNode=qListNode.next;
        }
        qListNode=pListNode.next;
        pListNode.next=qListNode.next;
        return head;
    }

结果：

leetcode_19_Remove Nth Node From End of List

标签：链表   移动   倒数第n个结点   

原文地址：http://blog.csdn.net/mnmlist/article/details/43446561