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.

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Linked List Two Pointers

/**
* 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 *temp = head;
int len=,m;
while(temp){
++len;
temp=temp->next;
}
if(n==len)
return head->next;
m=len-n;
ListNode *temp1=head;
while(m>){
temp1=temp1->next;
--m;
}
temp1->next=temp1->next->next;
return head; }
};

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