题解 | #删除链表的倒数第n个节点#
删除链表的倒数第n个节点
https://www.nowcoder.com/practice/f95dcdafbde44b22a6d741baf71653f6
方法一：循环
方法二：递归
思路，首相找出链表的深度，然后根据n所在的位置找出要踢出的数据
注意：
1引用传递的问题，方法中只是传递的引用，如果引用值发生变化，ok.如果引用被指向另一个引用则会丢失原有数据
所以，要么是对象类型，通过对象传递，值改变
import java.util.*;
import java.util.Stack;
import java.util.concurrent.atomic.*;
public class Solution {
/**
*
* @param head ListNode类
* @param n int整型
* @return ListNode类
*/
public ListNode removeNthFromEnd (ListNode head, int n) {
AtomicInteger lineDeep = new AtomicInteger(0);
dealHead(head, lineDeep);
ListNode up = new ListNode(0);
up.next = head;
dealHead2(head, up, lineDeep, n);
return up.next;
}
public void dealHead2(ListNode head, ListNode up, AtomicInteger lineDeep, Integer n) {
int get = lineDeep.get();
if (get != n) {
ListNode node = head.next;
up.next = head;
ListNode npNew = up.next;
if (null != node) {
lineDeep.decrementAndGet();
dealHead2(node, npNew, lineDeep, n);
}
} else {
if (null == head.next) {
up.next = null;
} else {
up.next = head.next;
}
}
}
public void dealHead(ListNode head, AtomicInteger lineDeep) {
ListNode next = head.next;
lineDeep.addAndGet(1);
if (null != next ) {
dealHead(next, lineDeep);
}
}
}
查看17道真题和解析
