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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.

 

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:

• 链表中节点数目在范围 [1, 104] 内
• 1 <= Node.val <= 106
• 1 <= m, n <= 1000

 

进阶: 你能通过 就地 修改链表的方式解决这个问题吗?

解法

方法一：模拟

按照题意模拟，遍历链表，每次遍历 $m$ 个节点，然后删除 $n$ 个节点，直到链表尾部。

时间复杂度 $O(n)$，空间复杂度 $O(1)$。

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 _ in range(m - 1):
                if pre:
                    pre = pre.next
            if pre is None:
                return head
            cur = pre
            for _ 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;
    }
}

C++

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

Go

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func deleteNodes(head *ListNode, m int, n int) *ListNode {
	pre := head
	for pre != nil {
		for i := 0; i < m-1 && pre != nil; i++ {
			pre = pre.Next
		}
		if pre == nil {
			return head
		}
		cur := pre
		for i := 0; i < n && cur != nil; i++ {
			cur = cur.Next
		}
		pre.Next = nil
		if cur != nil {
			pre.Next = cur.Next
		}
		pre = pre.Next
	}
	return head
}

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