main.go

// Source: https://leetcode.com/problems/partition-list
// Title: Partition List
// Difficulty: Medium
// Author: Mu Yang <http://muyang.pro>

////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
// 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
//
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////

package main

type ListNode struct {
	Val  int
	Next *ListNode
}

func partition(head *ListNode, x int) *ListNode {
	leftHead := &ListNode{}
	rightHead := &ListNode{}

	leftTail := leftHead
	rightTail := rightHead

	for head != nil {
		if head.Val < x {
			leftTail.Next = head
			leftTail = head
		} else {
			rightTail.Next = head
			rightTail = head
		}
		head = head.Next
	}

	leftTail.Next = rightHead.Next
	rightTail.Next = nil

	return leftHead.Next
}