Question

2074. Reverse Nodes in Even Length Groups

中文文档

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

Solution 1

# 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.
 * 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 {
 *   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;
}