【LeetCode OJ】Remove Nth Node From End of List

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题目：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.
struct ListNode {
	int val;
	ListNode *next;
	ListNode(int x) : val(x), next(NULL) {}
	
};
ListNode *removeNthFromEnd(ListNode *head, int n)
{
	ListNode *r = head, *s;
	if (head == NULL)
		return head;
	int i = 0;
	for (ListNode *p = head; p != NULL; p = p->next)
	{
		i++; 
	}
	if (i == n)  //删除第一个节点
	{
		ListNode *l = head->next;
		free(head);
		return l;
	}
	for (int num = 0; num < i - n - 1; num++)//找到要删节点的前一个节点
	{
		r = r->next;
	}
	ListNode *tmp = r->next;
	r->next = r->next->next;
	free(tmp);
	return head;
	
}

【LeetCode OJ】Remove Nth Node From End of List

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原文地址：http://blog.csdn.net/xujian_2014/article/details/44779593