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2074. Reverse Nodes in Even Length Groups

Description

You are given the head of a linked list.

The nodes in the linked list are sequentially assigned to non-empty groups whose lengths form the sequence of the natural numbers (1, 2, 3, 4, ...). The length of a group is the number of nodes assigned to it. In other words,

  • The 1st node is assigned to the first group.
  • The 2nd and the 3rd nodes are assigned to the second group.
  • The 4th, 5th, and 6th nodes are assigned to the third group, and so on.

Note that the length of the last group may be less than or equal to 1 + the length of the second to last group.

Reverse the nodes in each group with an even length, and return the head of the modified linked list.

 

Example 1:

Input: head = [5,2,6,3,9,1,7,3,8,4]
Output: [5,6,2,3,9,1,4,8,3,7]
Explanation:
- The length of the first group is 1, which is odd, hence no reversal occurs.
- The length of the second group is 2, which is even, hence the nodes are reversed.
- The length of the third group is 3, which is odd, hence no reversal occurs.
- The length of the last group is 4, which is even, hence the nodes are reversed.

Example 2:

Input: head = [1,1,0,6]
Output: [1,0,1,6]
Explanation:
- The length of the first group is 1. No reversal occurs.
- The length of the second group is 2. The nodes are reversed.
- The length of the last group is 1. No reversal occurs.

Example 3:

Input: head = [1,1,0,6,5]
Output: [1,0,1,5,6]
Explanation:
- The length of the first group is 1. No reversal occurs.
- The length of the second group is 2. The nodes are reversed.
- The length of the last group is 2. The nodes are reversed.

 

Constraints:

  • The number of nodes in the list is in the range [1, 105].
  • 0 <= Node.val <= 105

Solutions

  • /**
 * 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 reverseEvenLengthGroups(ListNode head) {
        int n = 0;
        for (ListNode t = head; t != null; t = t.next) {
            ++n;
        }
        ListNode dummy = new ListNode(0, head);
        ListNode prev = dummy;
        int l = 1;
        for (; (1 + l) * l / 2 <= n && prev != null; ++l) {
            if (l % 2 == 0) {
                ListNode node = prev.next;
                prev.next = reverse(node, l);
            }
            for (int i = 0; i < l && prev != null; ++i) {
                prev = prev.next;
            }
        }
        int left = n - l * (l - 1) / 2;
        if (left > 0 && left % 2 == 0) {
            ListNode node = prev.next;
            prev.next = reverse(node, left);
        }
        return dummy.next;
    }

    private ListNode reverse(ListNode head, int l) {
        ListNode prev = null;
        ListNode cur = head;
        ListNode tail = cur;
        int i = 0;
        while (cur != null && i < l) {
            ListNode t = cur.next;
            cur.next = prev;
            prev = cur;
            cur = t;
            ++i;
        }
        tail.next = cur;
        return prev;
    }
}

  • # Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def reverseEvenLengthGroups(self, head: Optional[ListNode]) -> Optional[ListNode]:
        def reverse(head, l):
            prev, cur, tail = None, head, head
            i = 0
            while cur and i < l:
                t = cur.next
                cur.next = prev
                prev = cur
                cur = t
                i += 1
            tail.next = cur
            return prev

        n = 0
        t = head
        while t:
            t = t.next
            n += 1
        dummy = ListNode(0, head)
        prev = dummy
        l = 1
        while (1 + l) * l // 2 <= n and prev:
            if l % 2 == 0:
                prev.next = reverse(prev.next, l)
            i = 0
            while i < l and prev:
                prev = prev.next
                i += 1
            l += 1
        left = n - l * (l - 1) // 2
        if left > 0 and left % 2 == 0:
            prev.next = reverse(prev.next, left)
        return dummy.next


  • /**
 * Definition for singly-linked list.
 * class ListNode {
 *     val: number
 *     next: ListNode | null
 *     constructor(val?: number, next?: ListNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.next = (next===undefined ? null : next)
 *     }
 * }
 */

function reverseEvenLengthGroups(head: ListNode | null): ListNode | null {
    let nums = [];
    let cur = head;
    while (cur) {
        nums.push(cur.val);
        cur = cur.next;
    }

    const n = nums.length;
    for (let i = 0, k = 1; i < n; i += k, k++) {
        // 最后一组， 可能出现不足
        k = Math.min(n - i, k);
        if (!(k & 1)) {
            let tmp = nums.splice(i, k);
            tmp.reverse();
            nums.splice(i, 0, ...tmp);
        }
    }

    cur = head;
    for (let num of nums) {
        cur.val = num;
        cur = cur.next;
    }
    return head;
}

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