leetcode19Remove 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:

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode * function(ListNode * head, int * n)
    {
        ListNode * result;
        
        if (head->next)
        {
            result = function(head->next, n);
        }
        else
        {
            result = NULL;
        }
        
        if (--*n == 0)
        {
            return result;
        }
        else
        {
            head->next = result;
            return head;
        }
    }
    
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        int m = n;
        if (head == NULL)
        {
            return NULL;
        }
        
        ListNode * result = function(head, &m);
        
        return result;
    }
};
 

class Solution
{
public:
    ListNode* removeNthFromEnd(ListNode* head, int n)
    {
        ListNode** t1 = &head, *t2 = head;
        for(int i = 1; i < n; ++i)
        {
            t2 = t2->next;
        }
        while(t2->next != NULL)
        {
            t1 = &((*t1)->next);
            t2 = t2->next;
        }
        *t1 = (*t1)->next;
        return head;
    }
};