LeetCode进阶之路（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.

题目：删除链表倒数第N个节点。

思路：准备两个指针P、Q，先让P走N个，然后在一起走，知道P走完，此时Q到达的点就是需要删除的点。

public static ListNode removeNthFromEnd(ListNode head, int n) {  
        if(n == 0 || head == null){  
            return head;  
        }  
        if(n == 1 && head.next==null){  
            return null;  
        }  
           
        ListNode p = head, q = head;  
        // 让p先行q n个位置  
        for(int i=0; i<n; i++){  
            if(p != null){  
                p = p.next;  
            }else{  
                return head;  
            }  
        }  
           
        // 如果这个时候p已经是null，则说明删除的必定为head  
        if(p == null){  
            head = head.next;  
            return head;  
        }  
           
        // p和q一起前进  
        while(p.next != null){  
            q = q.next;  
            p = p.next;  
        }  
        // 删除元素  
        q.next = q.next.next;  
        return head;  
    }  

种一棵树最好的时间是十年前，其次是现在！