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.

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */

 //题目要求一次遍历list即实现节点移除，使用双引用p与q来实现,q-p=n，向后遍历，当q无后继时到达list尾端，则p的next即为欲删除节点
public class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        if (head == null) return head;
        if (head.next == null) return null;    //list只有一个node，潜台词就是n=1
        
        ListNode newhead = new ListNode(Integer.MIN_VALUE);    //被删除的节点可能是head，故设置一个newhead
        newhead.next = head;
        ListNode p = head;                //利用双引用实现一次遍历即删除倒数节点的目的
        ListNode q = head;                //示例list：1 2 3 4 5  5即为n=1时的情况
        
        for(int i=0;i<n;i++) q=q.next;    //运行了n次next，此时q为第n+1个节点，若n=2则p=1,q=3
        
        if(q == null){
        newhead.next = head.next;         //如果此时q即为null，则n必为list.length，即欲删除的为head
        return newhead.next;              //!!注意，此时必须return，否则继续向下执行q.next会出现空引用异常
        } 
        
        while(q.next != null){            //使p与q向后遍历，直至使q为list尾节点
            p = p.next;
            q = q.next;
        }
        ListNode removenode = p.next;     //此时，p.next即为欲删除的节点
        p.next = removenode.next;
        
        return newhead.next;
    }
}