[LeetCode] 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.

解法：删除列表中倒数第n个节点，设置两个节点指针ptr1和ptr2,它们之间相差n个节点，然后将指针往下移动，当ptr2的下一个节点为NULL时，ptr1就指向了删除节点的前一个节点。

特殊情况测试用例：
[1],1
[1,2],1
[1,2],2

C语言实现：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 * int val;
 * struct ListNode *next;
 * };
 */
struct ListNode* removeNthFromEnd(struct ListNode* head, int n) {
    if(NULL == head )
        return NULL;
    struct ListNode *ptr1,*ptr2,*tmp = NULL;
    int i;
    ptr1 = head;
    ptr2 = head;
    for(i=0;i<n;i++)
    {
        ptr2 = ptr2->next;
    }
    if(NULL == ptr2)
    {
        tmp = head;
        head = head->next;
        free(tmp);
        return head;
    }
    while(ptr2->next)
    {
        ptr2 = ptr2->next;
        ptr1 = ptr1->next;
    }
    tmp = ptr1->next;
    ptr1->next = (ptr1->next)->next;
    free(tmp);
    return head;
}