[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.

解题思路：

One pass 要求感觉有些鸡肋~ = =

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *removeNthFromEnd(ListNode *head, int n) {
        ListNode *first = head, *second = head;
        if(first == NULL) return NULL;
        for(int i = 0;i < n;i++)
        {
            if(i == n - 1 && first -> next == NULL) return head -> next;
            else first = first -> next;
        }
        while(first -> next != NULL)
        {
            first = first -> next;
            second = second -> next;
        }
        
        ListNode *tmp = second -> next;
        second -> next = tmp -> next;
        delete tmp;
        return head;
    }
};