【leetcode】19. Remove Nth From End of List(C)

Description：

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

Example1:

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.

[ 题目链接 ]

提交代码：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* removeNthFromEnd(struct ListNode* head, int n) 
{
	if (head == NULL||head->next==NULL)
		return NULL;

	int i;
	int len=0,pos;
	struct ListNode* p = head;

	//calculate the len of the list
	while (p != NULL)
	{
		len++;
		p = p->next;
	}

	pos = len - n;
    p = head;
    if (pos == 0)
        return p->next;
	
	
	for (i = 0; i < pos-1; i++)
		p = p->next;

	struct ListNode* tmp = p->next;
	p->next = p->next->next;
	free(tmp);
	return head;
}

运行结果：

其他：
提交历史中有这个报错：

报错的代码行在这里：

报错的原因是刚开始我忘记加[p=head;]这句代码了，所以p的指向是混乱的，
所以我后面return p->next的时候，编译器就不知道我的p到底指向哪里。