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Leetcode 线性表 Remove Nth Node From End of List

时间:2014-05-15 02:53:13      阅读:241      评论:0      收藏:0      [点我收藏+]

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本文为senlie原创,转载请保留此地址:http://blog.csdn.net/zhengsenlie


Remove Nth Node From End of List

 Total Accepted: 12160 Total Submissions: 41280

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.



题意:移除链表的倒数第n个元素
思路:
两个指针p, q,
p先走n步,然后p,q一起走,当p走到尾的时候,q->next就是要删除的节点
复杂度: 时间O(n),空间O(1)


/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *removeNthFromEnd(ListNode *head, int n) {
    	ListNode dummy(0);
    	dummy.next = head;
    	ListNode *p, *q;
    	p = q = &dummy;
    	while(n--) p= p->next;
    	while(p->next){
    		p = p->next;
    		q = q->next;
    	}
    	ListNode *to_delete = q->next;
    	q->next = to_delete->next;
    	delete to_delete;
    	return dummy.next;
    }
};


Leetcode 线性表 Remove Nth Node From End of List,布布扣,bubuko.com

Leetcode 线性表 Remove Nth Node From End of List

标签:style   blog   class   code   c   tar   

原文地址:http://blog.csdn.net/zhengsenlie/article/details/25797127

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