Remove Nth Node from End of List

Given A linked list, remove the nth node from the end of the 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 would always be valid.
Try to do the in one pass.

The usual solution is to rotation the total length of the list, calculate the index (LEN-N) that needs to be deleted, but it takes two cycles. The code is as follows:

classSolution { Public: ListNode* Removenthfromend (listnode* head,intN) {intCNT =0; ListNode* node =Head;  while(node) {++CNT; Node= node->Next; }         intindex = CNT-N; if(0==index) {Head= head->Next; } Else{node=Head;  while(--index) {Node= node->Next; } node->next = node->next->Next; }        returnHead; }};

In the solution above, you have to calculate the Len before you can calculate the index. We can use another method to calculate index, first loop n times, so that node self-increment. A new pointer slow to head, then node and slow increment, node is empty, slow is just the point to delete. The code is as follows:

int N) {    **slow = &head, *fast = head;          while (--n) fast = Fast->next;      while (fast->Next) {        = & ((*slow),next);         = Fast->next;    }         *slow = (*slow),next;         return head;}

Remove Nth Node from End of List