19删除链表倒数第N个节点
删除链表中倒数第N个节点
给你一个链表,删除链表的倒数第 n
个结点,并且返回链表的头结点。
1、最容易想到的,把删除倒数第N个节点转换成删除正数第M个节点
class Solution {
public ListNode removeNthFromEnd(ListNode head, int n) {
if(head == null) return null;
ListNode header = new ListNode(-1);
header.next = head;
//找出链表长度,问题转换成删除链表第n个节点
int len = 1;
ListNode temp = head;
while(temp.next != null){
len++;
temp = temp.next;
}
int target = len - n + 1;
ListNode p = header;
for(int i = 0;i < target - 1;i++,p = p.next);
p.next = p.next.next;
//System.out.print(p.val);
return header.next;
}
}
2、快慢指针
添加一个虚拟头节点header,初始是快慢指针fast、low = header,记录快慢指针之间的距离,快指针先走n步后,慢指针开始走,当fast.next == null 时,fast.next即为倒数第n个节点,删除即可
class Solution {
public ListNode removeNthFromEnd(ListNode head, int n) {
ListNode header = new ListNode(-1);
header.next = head;
ListNode fast = header;
ListNode slow = header;
int i = 0;//记录快慢指针的距离
while(fast.next != null){
if(i >= n){
slow = slow.next;
}
i++;
fast = fast.next;
}
slow.next = slow.next.next;
return header.next;
}
}