LeetCode in TypeScript
86. Partition List
Medium
Given the head of a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x.
You should preserve the original relative order of the nodes in each of the two partitions.
Example 1:
Input: head = [1,4,3,2,5,2], x = 3
Output: [1,2,2,4,3,5]
Example 2:
Input: head = [2,1], x = 2
Output: [1,2]
Constraints:
- The number of nodes in the list is in the range
[0, 200]. -100 <= Node.val <= 100-200 <= x <= 200
Solution
import { ListNode } from '../../com_github_leetcode/listnode'
/**
* 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 partition(head: ListNode | null, x: number): ListNode | null {
let beforeHead = new ListNode(0)
let afterHead = new ListNode(0)
let before = beforeHead
let after = afterHead
while (head !== null) {
const nextNode = head.next
if (head.val < x) {
before.next = head
before = before.next
} else {
after.next = head
after = after.next
}
head = nextNode
}
after.next = null
before.next = afterHead.next
return beforeHead.next
}
export { partition }