Description:
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.
Solution:
Time Complexity : O(n)
Space Complexity: O(1)
var oddEvenList = function(head) {
// Handle base cases
if(!head || !head.next || !head.next.next) {
return head
}
// Set two pointers
let cur = head
let next = head.next
// Set the odd.next to point to the next even node
// Move each pointer up one node on each iteration
while(next && next.next) {
const temp = next.next
const temp2 = cur.next
cur.next = temp
next.next = temp.next
temp.next = temp2
cur = cur.next
next = next.next
}
return head
};
Top comments (1)
Can't write code on the phone...
ArrOdd = arrOriginal.filter(i, idx) => idx % 2 == 0
ArrEven = arrOriginal.filter(i, idx) => idx % 2 == 1
Return (arrOdd...arrEven)