[LeetCode] 1474. Delete N Nodes After M Nodes of a Linked List

Given the head of a linked list and two integers m and n. Traverse the linked list and remove some nodes in the following way:

• Start with the head as the current node.
• Keep the first m nodes starting with the current node.
• Remove the next n nodes
• Keep repeating steps 2 and 3 until you reach the end of the list.

Return the head of the modified list after removing the mentioned nodes.

Follow up question: How can you solve this problem by modifying the list in-place?

Example 1:

Input: head = [1,2,3,4,5,6,7,8,9,10,11,12,13], m = 2, n = 3
Output: [1,2,6,7,11,12]
Explanation: Keep the first (m = 2) nodes starting from the head of the linked List  (1 ->2) show in black nodes.
Delete the next (n = 3) nodes (3 -> 4 -> 5) show in read nodes.
Continue with the same procedure until reaching the tail of the Linked List.
Head of linked list after removing nodes is returned.

Example 2:

Input: head = [1,2,3,4,5,6,7,8,9,10,11], m = 1, n = 3
Output: [1,5,9]
Explanation: Head of linked list after removing nodes is returned.

Example 3:

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

Example 4:

Input: head = [9,3,7,7,9,10,8,2], m = 1, n = 2
Output: [9,7,8]

Constraints:

• The given linked list will contain between 1 and 10^4 nodes.
• The value of each node in the linked list will be in the range [1, 10^6].
• 1 <= m,n <= 1000

删除链表 M 个节点之后的 N 个节点。

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?

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/delete-n-nodes-after-m-nodes-of-a-linked-list
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这是一道链表题。链表题如果在面试中出现，属于送分题，一定要会。

题意不难懂，对于给定的 input 链表，我们先保留 m 个节点，再跳过 n 个节点，以此交替，直到遍历完链表。那么做法也是很直接，我们给两个变量 i 和 j 分别去追踪到底数了几个节点了。这里我们还需要一个 pre 节点，记录需要跳过的 n 个节点之前的一个节点，这样在跳过 n 个节点之后，我们可以把跳过部分之前的和之后的节点连在一起。代码应该很好理解。

时间O(n)

空间O(1)

Java实现

class Solution {
    public ListNode deleteNodes(ListNode head, int m, int n) {
        ListNode cur = head;
        ListNode pre = null;
        while (cur != null) {
            int i = m;
            int j = n;
            while (cur != null && i > 0) {
                pre = cur;
                cur = cur.next;
                i--;
            }
            while (cur != null && j > 0) {
                cur = cur.next;
                j--;
            }
            pre.next = cur;
        }
        return head;
    }
}

 

LeetCode 题目总结