leetcode (19) - Remove Nth Node From End of List

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* removeNthFromEnd(struct ListNode* head, int n) {
        if(head == NULL || n <= 0)
            return head;
        
        struct ListNode* first = head;
        struct ListNode* second = head;
        struct ListNode* pre = NULL;
        
        for(int i = 0; i < n - 1; i++) // 中间间隔N-1的距离
            first = first->next;
            
        while(first->next)
        {
            pre = second;  // pre是倒数第k个节点的前一个节点，用来删除
            second = second->next;  // second是倒数第k个节点
            first = first->next;
        }
        
        if( !pre )  // 如果倒数第k个节点就是head，那么pre会保留原来的NULL，这时head直接是倒数第k个节点second的下一个节点
            head = second->next;
        else
            pre->next = second->next;
        
        return head;
}