25. Reverse Nodes in k-Group

Description

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:

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

Example 2:

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?

Solutions

Solution 1: Iteration

The time complexity is $O(n)$, and the space complexity is $O(1)$. Here, $n$ is the length of the linked list.

Python3JavaGoTypeScriptRustC#PHP

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def reverseKGroup(self, head: ListNode, k: int) -> ListNode:
        def reverseList(head):
            pre, p = None, head
            while p:
                q = p.next
                p.next = pre
                pre = p
                p = q
            return pre

        dummy = ListNode(next=head)
        pre = cur = dummy
        while cur.next:
            for _ in range(k):
                cur = cur.next
                if cur is None:
                    return dummy.next
            t = cur.next
            cur.next = None
            start = pre.next
            pre.next = reverseList(start)
            start.next = t
            pre = start
            cur = pre
        return dummy.next

/**
 * 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 reverseKGroup(ListNode head, int k) {
        ListNode dummy = new ListNode(0, head);
        ListNode pre = dummy, cur = dummy;
        while (cur.next != null) {
            for (int i = 0; i < k && cur != null; ++i) {
                cur = cur.next;
            }
            if (cur == null) {
                return dummy.next;
            }
            ListNode t = cur.next;
            cur.next = null;
            ListNode start = pre.next;
            pre.next = reverseList(start);
            start.next = t;
            pre = start;
            cur = pre;
        }
        return dummy.next;
    }

    private ListNode reverseList(ListNode head) {
        ListNode pre = null, p = head;
        while (p != null) {
            ListNode q = p.next;
            p.next = pre;
            pre = p;
            p = q;
        }
        return pre;
    }
}

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func reverseKGroup(head *ListNode, k int) *ListNode {
    var dummy *ListNode = &ListNode{}
    p, cur := dummy, head
    for cur != nil {
        start := cur
        for i := 0; i < k; i++ {
            if cur == nil {
                p.Next = start
                return dummy.Next
            }
            cur = cur.Next
        }
        p.Next, p = reverse(start, cur), start
    }
    return dummy.Next
}

func reverse(start, end *ListNode) *ListNode {
    var pre *ListNode = nil
    for start != end {
        tmp := start.Next
        start.Next, pre = pre, start
        start = tmp
    }
    return pre
}