25. Reverse Nodes in k-Group

  • Hardness: \(\color{red}\textsf{Hard}\)
  • Ralated Topics: Linked List、Recursion

一、題目

Given the head of a linked list, reverse the nodes of the list k at a time, and return the 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.
You may not alter the values in the list’s nodes, only nodes themselves may be changed.

Example 1:
reverse_ex1

  • Input: head = [1,2,3,4,5], k = 2
  • Output: [2,1,4,3,5]

Example 2: reverse_ex2

  • Input: head = [1,2,3,4,5], k = 3
  • Output: [3,2,1,4,5]

Constraints:

  • The number of nodes in the list is n.
  • 1 <= k <= n <= 5000
  • 0 <= Node.val <= 1000

二、分析

  • 經典的 ListNode 問題，鏈表的後序遍歷，利用後序遍歷回傳值的特性並用 recursion 來完成這一題。

三、解題

1. Recursion

  • Time complexity: \(O()\)
  • Space complexity: \(O()\)
ListNode* reverseKGroup(ListNode* head, int k) {
    int cnt = k;
    ListNode* last = head;
    while (cnt && last) {
        last = last->next;
        cnt--;
    }
    if (cnt == 0) {
        last = reverseKGroup(last, k);
        ListNode* prev = nullptr;
        ListNode* curr = head;
        ListNode* next = nullptr;
        cnt = k;
        while (cnt--) {
            next = curr->next;
            curr->next = prev;
            prev = curr;
            curr = next;
        }
        head->next = last;
        head = prev;
    }
    return head;
}

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