Leetcode: 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.

分析：two pointers.

class Solution {
public:
    ListNode *removeNthFromEnd(ListNode *head, int n) {
        if(head == NULL) return head;
        
        ListNode *dummy = new ListNode(-1);
        dummy->next = head;
        
　　　　　//move fast pointer to (n+1)th node from the head
        ListNode *fast = head;
        for(int i = 0; i < n; i++)
            fast = fast->next;
        //move fast to the end of the list
        ListNode *prev = dummy, *p = head;
        while(fast){
            fast = fast->next;
            prev = p;
            p = p->next;
        }
        prev->next = p->next;
        
        return dummy->next;
    }
};