[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.

思路：两个指针 p, q，其中q先走n步，之后如果q为null，则删除第一个node，否则，p, q开始一起走，直到q.next为null，此时p.next即是要删除的node.

JAVA代码：

public ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode p = head;
        ListNode q = head;
        int i = 0;
        while (i < n) {
            q = q.next;
            i ++;
        }
        if (q == null) {
            head = head.next;
            return head;
        }
        while (q.next != null) {
            q = q.next;
            p = p.next;
        }
        p.next = p.next.next;
        return head;    
    }