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.

Follow up:

  • Could you solve the problem in O(1) extra memory space?
  • You may not alter the values in the list's nodes, only nodes itself 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]

Example 3:

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

Example 4:

Input: head = [1], k = 1
Output: [1]

 

Constraints:

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

Rust Solution

struct Solution;
use rustgym_util::*;
use std::collections::VecDeque;

impl Solution {
    fn reverse_k_group(head: ListLink, k: i32) -> ListLink {
        let mut p = head;
        let mut count = 0;
        let mut queue: VecDeque<ListLink> = VecDeque::new();
        let k = k as usize;
        while let Some(mut node) = p {
            p = node.next.take();
            queue.push_back(Some(node));
            count += 1;
            if count == k {
                break;
            }
        }
        if queue.len() == k {
            let mut prev: ListLink = Self::reverse_k_group(p, k as i32);
            while let Some(link) = queue.pop_front() {
                if let Some(mut node) = link {
                    node.next = prev;
                    prev = Some(node);
                }
            }
            prev
        } else {
            let mut prev: ListLink = None;
            while let Some(link) = queue.pop_back() {
                if let Some(mut node) = link {
                    node.next = prev;
                    prev = Some(node);
                }
            }
            prev
        }
    }
}

#[test]
fn test() {
    let head = list!(1, 2, 3, 4, 5);
    let k = 2;
    let res = list!(2, 1, 4, 3, 5);
    assert_eq!(Solution::reverse_k_group(head, k), res);
    let head = list!(1, 2, 3, 4, 5);
    let k = 3;
    let res = list!(3, 2, 1, 4, 5);
    assert_eq!(Solution::reverse_k_group(head, k), res);
    let head = list!(1, 2);
    let k = 2;
    let res = list!(2, 1);
    assert_eq!(Solution::reverse_k_group(head, k), res);
}

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