LeetCode Remove Nth Node From End of List 删除倒数第n个元素

Remove Nth Node From End of List

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.
这里的Note的意思应该就是可以不做特殊处理n值了。不过其实特殊处理也就一句语句。

思路：

快指针和慢指针的概念，先一个指针先走，到了离起点n个节点远的时候，慢指针走，等到快指针走到结尾，那么慢指针走到了需要删除的位置了。

初学者注意：保留慢指针的前置指针，这样才能删除。

注意：特殊处理需要删除头指针的情况，如输入1->2->3->4; n=4，那么需要删除1，这就需要特殊处理了，头指针往前一步就可以了。

struct ListNode {
	int val;
	ListNode *next;
	ListNode(int x) : val(x), next(NULL) {}
};

class Solution {
public:
	ListNode *removeNthFromEnd(ListNode *head, int n)
	{
		ListNode *end = head;
		int i = 0;
		for (; i < n; i++)
		{
			if (end) end = end->next;
			else break;
		}
		if (i != n) return head;

		findDeletePos(head, head, head, end);
		return head;

	}

	void findDeletePos(ListNode *(&head), ListNode *pre, ListNode *cur, ListNode *end)
	{
		while (end)
		{
			end = end->next;
			pre = cur;
			cur = cur->next;
		}
		//注意：删除头指针的时候要做特殊处理
		if (pre == cur)
		{
			head = head->next;
			return;
		}
		pre->next = cur->next;
	}
};

//2015-1-25 update

ListNode *removeNthFromEnd(ListNode *head, int n)
	{
		ListNode dummy(-1);
		dummy.next = head;
		ListNode *pre = &dummy;

		while (--n > 0) head = head->next;

		while (head->next)
		{
			pre = pre->next;
			head = head->next;
		}
		pre->next = pre->next->next;
		return dummy.next;
	}