Published May 27, 2022
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Given the head of a singly linked list, group all the nodes with odd indices
together followed by the nodes with even indices, and return the reordered
list.
The first node is considered odd, and the second node is even, and so on.
Note that the relative order inside both the even and odd groups should remain as it was in the input.
You must solve the problem in O(1) extra space complexity and O(n) time complexity.
Input: head = [1,2,3,4,5]
Output: [1,3,5,2,4]
Input: head = [2,1,3,5,6,4,7]
Output: [2,3,6,7,1,5,4]
[0, 10^4].-10^6 <= Node.val <= 10^6/**
* 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 oddEvenList(head: ListNode | null): ListNode | null {
if(head === null || head.next === null || head.next.next === null){
return head;
}
let h1: ListNode = head;
let h2: ListNode = head.next;
const dummy: ListNode = h2;
while(h2 && h2.next){
h1.next = h2.next;
h1 = h1.next;
h2.next = h1.next
h2 = h2.next;
}
h1.next = dummy;
return head;
};