[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

刪除鏈表 M 個節點之后的 N 個節點。

給定鏈表 head 和兩個整數 m 和 n. 遍歷該鏈表並按照如下方式刪除節點:

開始時以頭節點作為當前節點.
保留以當前節點開始的前 m 個節點.
刪除接下來的 n 個節點.
重復步驟 2 和 3, 直到到達鏈表結尾.
在刪除了指定結點之后, 返回修改過后的鏈表的頭節點.

進階問題: 你能通過就地修改鏈表的方式解決這個問題嗎?

來源：力扣（LeetCode）
鏈接：https://leetcode-cn.com/problems/delete-n-nodes-after-m-nodes-of-a-linked-list
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這是一道鏈表題。鏈表題如果在面試中出現，屬於送分題，一定要會。

題意不難懂，對於給定的 input 鏈表，我們先保留 m 個節點，再跳過 n 個節點，以此交替，直到遍歷完鏈表。那么做法也是很直接，我們給兩個變量 i 和 j 分別去追蹤到底數了幾個節點了。這里我們還需要一個 pre 節點，記錄需要跳過的 n 個節點之前的一個節點，這樣在跳過 n 個節點之后，我們可以把跳過部分之前的和之后的節點連在一起。代碼應該很好理解。

時間O(n)

空間O(1)

Java實現

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;
    }
}

 

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