# 86. Partition List

## Description

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

## Solutions

Solution 1: Simulation

We create two linked lists, one to store nodes less than $x$, and the other to store nodes greater than or equal to $x$. Then we concatenate them.

The time complexity is $O(n)$, where $n$ is the length of the original linked list. The space complexity is $O(1)$.

• /**
* 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 partition(ListNode head, int x) {
ListNode d1 = new ListNode();
ListNode d2 = new ListNode();
ListNode t1 = d1, t2 = d2;
t1 = t1.next;
} else {
t2 = t2.next;
}
}
t1.next = d2.next;
t2.next = null;
return d1.next;
}
}

• /**
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode() : val(0), next(nullptr) {}
*     ListNode(int x) : val(x), next(nullptr) {}
*     ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
ListNode* partition(ListNode* head, int x) {
ListNode* d1 = new ListNode();
ListNode* d2 = new ListNode();
ListNode* t1 = d1;
ListNode* t2 = d2;
t1 = t1->next;
} else {
t2 = t2->next;
}
}
t1->next = d2->next;
t2->next = nullptr;
return d1->next;
}
};

• # Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
def partition(self, head: Optional[ListNode], x: int) -> Optional[ListNode]:
d1, d2 = ListNode(), ListNode()
t1, t2 = d1, d2
t1 = t1.next
else:
t2 = t2.next
t1.next = d2.next
t2.next = None
return d1.next


• /**
* type ListNode struct {
*     Val int
*     Next *ListNode
* }
*/
func partition(head *ListNode, x int) *ListNode {
d1, d2 := &ListNode{}, &ListNode{}
t1, t2 := d1, d2
t1 = t1.Next
} else {
t2 = t2.Next
}
}
t1.Next = d2.Next
t2.Next = nil
return d1.Next
}

• /**
* function ListNode(val, next) {
*     this.val = (val===undefined ? 0 : val)
*     this.next = (next===undefined ? null : next)
* }
*/
/**
* @param {number} x
* @return {ListNode}
*/
var partition = function (head, x) {
const d1 = new ListNode();
const d2 = new ListNode();
let t1 = d1,
t2 = d2;
t1 = t1.next;
} else {
t2 = t2.next;
}
}
t1.next = d2.next;
t2.next = null;
return d1.next;
};


• // Definition for singly-linked list.
// #[derive(PartialEq, Eq, Clone, Debug)]
// pub struct ListNode {
//   pub val: i32,
//   pub next: Option<Box<ListNode>>
// }
//
// impl ListNode {
//   #[inline]
//   fn new(val: i32) -> Self {
//     ListNode {
//       next: None,
//       val
//     }
//   }
// }
impl Solution {
pub fn partition(head: Option<Box<ListNode>>, x: i32) -> Option<Box<ListNode>> {
let mut d1 = Some(Box::new(ListNode::new(0)));
let mut d2 = Some(Box::new(ListNode::new(0)));
let (mut t1, mut t2) = (&mut d1, &mut d2);
while let Some(mut node) = head {
if node.val < x {
t1.as_mut().unwrap().next = Some(node);
t1 = &mut t1.as_mut().unwrap().next;
} else {
t2.as_mut().unwrap().next = Some(node);
t2 = &mut t2.as_mut().unwrap().next;
}
}
t1.as_mut().unwrap().next = d2.unwrap().next;
d1.unwrap().next
}
}