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;
}
}