leetcode.19----------Remove Nth Node From End of List

标签：remove nth node from   算法   leetcode   acm   

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.

Runtime: 9 ms

class Solution{
public:
	ListNode *removeNthFromEnd(ListNode *head, int n)
	{
		if (head == NULL)
			return NULL;

		ListNode*p1 = head;
		ListNode*p2 = head;
		ListNode*pre = NULL;
		while (n-1 != 0)
		{
			p1 = p1->next;
			n--;
		}

		while (p1->next)
		{
			pre = p2;
			p2 = p2->next;
			p1 = p1->next;
		}

		if (p2 == head)
			head = head->next;
		else
		{
			pre->next = p2->next;
			delete p2;
		}
		return head;
	}
};

leetcode.19----------Remove Nth Node From End of List

标签：remove nth node from   算法   leetcode   acm   

原文地址：http://blog.csdn.net/chenxun_2010/article/details/43460647