LeetCode 2074. Reverse Nodes in Even Length Groups Solution in Java, Python & TypeScript | Explanation + Code

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

PythonJavaTypeScript

# 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(code-box)

/** * 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; } }(code-box)

/** * 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; }(code-box)