Welcome to Subscribe On Youtube
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 nonempty 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
1^{st}
node is assigned to the first group.  The
2^{nd}
and the3^{rd}
nodes are assigned to the second group.  The
4^{th}
,5^{th}
, and6^{th}
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, 10^{5}]
. 0 <= Node.val <= 10^{5}
Solutions

/** * Definition for singlylinked 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 singlylinked 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 singlylinked 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; }