LeetCode-in-Rust

25. Reverse Nodes in k-Group

Hard

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.

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]

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

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

Solution

// Definition for singly-linked list.
// pub struct ListNode {
//   pub val: i32,
//   pub next: Option<Box<ListNode>>
// }
// 
// impl ListNode {
//   #[inline]
//   fn new(val: i32) -> Self {
//     ListNode {
//       next: None,
//       val
//     }
//   }
// }
impl Solution {
    pub fn reverse_k_group(mut head: Option<Box<ListNode>>, k: i32) -> Option<Box<ListNode>> {
        Solution::reverse(head, k)
    }
    pub fn reverse(mut head:Option<Box<ListNode>>, k: i32) -> Option<Box<ListNode>> {
        if head.is_none() {
            return None;
        }
        let mut new_head: Option<Box<ListNode>> = None;
        let mut i = 0;
        while let Some(mut head_val) = head.take() {
            head = head_val.next.take();
            head_val.next = new_head;
            new_head = Some(head_val);
            i += 1;
            if i == k {
                break;
            }
        }
        if i != k {
            // have to back out and return original list
            return Solution::reverse(new_head, i);
        }
        // we now have two lists:
        // head -> rest of list
        // new_head -> reversed list
        // i cannot figure out how in rust you do this in one pass, 
        // so we will go walk the list again to get the tail of new_head and make it head
        // 'append'
        head = Solution::reverse(head, k);
        let mut tailw = &mut new_head;
        if let Some(ref mut tail_val) = tailw {
            let mut tail = tail_val;
            while let Some(ref mut tail_next) = tail.next {
                tail = tail_next;
            }
            tail.next = head;
        }
        new_head
    }
}

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