README.md

1474. 删除链表 M 个节点之后的 N 个节点

English Version

题目描述

给定链表 head 和两个整数 m 和 n. 遍历该链表并按照如下方式删除节点:

• 开始时以头节点作为当前节点.
• 保留以当前节点开始的前 m 个节点.
• 删除接下来的 n 个节点.
• 重复步骤 2 和 3, 直到到达链表结尾.

在删除了指定结点之后, 返回修改过后的链表的头节点.

进阶问题: 你能通过就地修改链表的方式解决这个问题吗?

 

示例 1:

输入: head = [1,2,3,4,5,6,7,8,9,10,11,12,13], m = 2, n = 3
输出: [1,2,6,7,11,12]
解析: 保留前(m = 2)个结点,  也就是以黑色节点表示的从链表头结点开始的结点(1 ->2).
删除接下来的(n = 3)个结点(3 -> 4 -> 5), 在图中以红色结点表示.
继续相同的操作, 直到链表的末尾.
返回删除结点之后的链表的头结点.

示例 2:

输入: head = [1,2,3,4,5,6,7,8,9,10,11], m = 1, n = 3
输出: [1,5,9]
解析: 返回删除结点之后的链表的头结点.

示例 3:

输入: head = [1,2,3,4,5,6,7,8,9,10,11], m = 3, n = 1
输出: [1,2,3,5,6,7,9,10,11]

示例 4:

输入: head = [9,3,7,7,9,10,8,2], m = 1, n = 2
输出: [9,7,8]

 

提示:

•  1 <= 链表结点数 <= 10^4
• [1 <= 链表的每一个结点值 <=10^6]
• 1 <= m,n <= 1000

解法

遍历链表，修改指针指向即可。

Python3

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def deleteNodes(self, head: ListNode, m: int, n: int) -> ListNode:
        pre = head
        while pre:
            for i in range(m - 1):
                if pre:
                    pre = pre.next
            if pre is None:
                return head
            cur = pre
            for i in range(n):
                if cur:
                    cur = cur.next
            pre.next = None if cur is None else cur.next
            pre = pre.next
        return head

Java

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode deleteNodes(ListNode head, int m, int n) {
        ListNode pre = head;
        while (pre != null) {
            for (int i = 0; i < m - 1 && pre != null; ++i) {
                pre = pre.next;
            }
            if (pre == null) {
                return head;
            }
            ListNode cur = pre;
            for (int i = 0; i < n && cur != null; ++i) {
                cur = cur.next;
            }
            pre.next = cur == null ? null : cur.next;
            pre = pre.next;
        }
        return head;
    }
}

...