LeetCode 25. Reverse Nodes in k-Group - Array

題目鏈接：25. Reverse Nodes in k-Group

Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.

k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.

Example:

Given this linked list: 1->2->3->4->5
For k = 2, you should return: 2->1->4->3->5
For k = 3, you should return: 3->2->1->4->5

Note:

• Only constant extra memory is allowed.
• You may not alter the values in the list’s nodes, only nodes itself may be changed.

題解

力扣將此題設置爲困難。不過我覺得這道題目並不難。

首先判斷是否爲空。
不爲空的話，用數組保存不超過 k 個數，如果有 k 個數，就逆序賦值到這 k 個結點，不足 k 個數直接退出。
返回原頭結點 head 即可。

時間複雜度：$O(2n)$。
空間複雜度：一個一維數組，$O(k)$。

Java代碼

/**
 * 2020-1-31 18:37:14
 */
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        if (head == null) {
            return null;
        }
        int[] arr = new int[k];// 保存k個數
        ListNode hk = head;// 每k個結點的頭結點
        while (hk != null) {// 每一段的頭結點是否存在
            ListNode p = hk;// 移動指針
            int len = 0;// 當前這一段有幾個結點
            while (len < k && p != null) {// 找不超過k個結點直到爲空
                arr[len] = p.val;// 保存結點的值
                ++len;// 個數加1
                p = p.next;// 向後移動指針
            }
            if (len < k) {
                break;// 不足k個結點，不用逆序，直接退出
            }
            for (int i = len - 1; i >= 0; --i) {// 有k個結點，逆序賦值
                hk.val = arr[i];
                hk = hk.next;
            }
        }
        return head;
    }
}

原文鏈接：https://blog.csdn.net/pfdvnah/article/details/104126675

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