leetcode

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.

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