25. Reverse Nodes in k-Group

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25. Reverse Nodes in k-Group

Description

Difficulty: Hard

Related Topics: Linked List, Recursion

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:

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

Example 2:

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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

Follow-up: Can you solve the problem in O(1) extra memory space?

Hints/Notes

  • 2023/08/03
  • Check if there are k nodes, otherwise just return head
  • Reverse(head, tail), then apply the function to further nodes
  • 0x3F’s solution(checked)
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class Solution {
public:
    ListNode* reverseKGroup(ListNode* head, int k) {
        return nullptr;
    }
}

Solution

Language: C++

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/**
 * 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* reverseKGroup(ListNode* head, int k) {
        ListNode *tail = head;
        for (int i = 0; i < k; i++) {
            if (tail) {
                tail = tail->next;
            } else {
                return head;
            }
        }
        ListNode *newHead = reverse(head, tail);
        head->next = reverseKGroup(tail, k);
        // Note:
        // At first i was trying to do reverseKGroup() first then reverse(), i.e
        //
        // reverseKGroup(tail, k);
        // ListNode *newHead = reverse(head, tail);
        //
        // But it run into problem because the tail has been changed and the afterwards
        // reverse would fail. It's also more complicated to understand than the current
        // implementation, i.e. reverse just return a reversed linked list ending with nullptr
        return newHead;
    }

    ListNode* reverse(ListNode* head, ListNode* tail) {
        ListNode *cur = head, *pre = nullptr, *tmp = nullptr;
        while (cur != tail) {
            tmp = cur->next;
            cur->next = pre;
            pre = cur;
            cur = tmp;
        }
        return pre;
    }
};

iterative:

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/**
 * 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* reverseKGroup(ListNode* head, int k) {
        ListNode* p = head;
        int n = 0;
        while (p) {
            p = p->next;
            n++;
        }
        ListNode dummy(0);
        dummy.next = head;
        ListNode *prev = nullptr, *cur = head, *p0 = &dummy;
        for (; n >= k; n -= k) {
            for (int i = 0; i < k; i++) {
                ListNode* nxt = cur->next;
                cur->next = prev;
                prev = cur;
                cur = nxt;
            }
            ListNode* nxt = p0->next;
            p0->next->next = cur;
            p0->next = prev;
            p0 = nxt;
        }
        return dummy.next;
    }
};