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.

?

class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        if (head==NULL || n<=0) {
            return NULL;
        }
        ListNode* pre = NULL;
        ListNode* pHead  = head;
        ListNode* pBehind  = head;
        for (int i = 0; i < n-1; i++) {
            if (pHead->next != NULL) {
                pHead = pHead->next;
            } else {
                return NULL;
            }
        }
        while (pHead->next != NULL) {
            pre = pBehind;
            pHead = pHead->next;
            pBehind = pBehind->next;
        }
        if (pre == NULL) {
            head = pBehind->next;
            delete pBehind;
        } else {
            pre->next = pBehind->next;
            delete pBehind;
        }
        return head;
    }
};

?

Remove Nth Node From End of List

原文：http://hcx2013.iteye.com/blog/2215864