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;
}
};