[LeetCode19] Remove Nth Node from End of List

Topic:

Given A linked list, remove the nth node from the end of the list and return its head.

For example,

   n = 2.   After removing the second node from the end, the linked list becomes 1->2->3->5.

Given a single linked list, remove the last nth node and return the head of the linked list

idea: maintain two pointers slow,fast,fast first move n step, then slow, fast move at the same time, until fast is empty, delete the node after slow can

Code:

1 /**2 * Definition for singly-linked list.3 * struct ListNode {4 * int val;5 * ListNode *next;6 * ListNode (int x): Val (x), Next (NULL) {}7  * };8  */9 classSolution {Ten  Public: Onelistnode* Removenthfromend (listnode* head,intN) { AListNode *prehead =NewListNode (0); -ListNode *fast = prehead, *slow =Prehead; -Slow->next =head; the          for(inti =0; I <= N; ++i) -Fast = Fast->Next; -          while(FAST) -         { +Fast = Fast->Next; -slow = slow->Next; +         } ASlow->next = slow->next->Next; at         returnPrehead->Next; -     } -};

[LeetCode19] Remove Nth Node from End of List