Welcome to Subscribe On Youtube

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 $O(1)$.

  • /**
 * Definition for singly-linked list.
 * 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;
        while (head != null) {
            if (head.val < x) {
                t1.next = head;
                t1 = t1.next;
            } else {
                t2.next = head;
                t2 = t2.next;
            }
            head = head.next;
        }
        t1.next = d2.next;
        t2.next = null;
        return d1.next;
    }
}

  • /**
 * Definition for singly-linked list.
 * 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;
        while (head) {
            if (head->val < x) {
                t1->next = head;
                t1 = t1->next;
            } else {
                t2->next = head;
                t2 = t2->next;
            }
            head = head->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
        while head:
            if head.val < x:
                t1.next = head
                t1 = t1.next
            else:
                t2.next = head
                t2 = t2.next
            head = head.next
        t1.next = d2.next
        t2.next = None
        return d1.next


  • /**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func partition(head *ListNode, x int) *ListNode {
	d1, d2 := &ListNode{}, &ListNode{}
	t1, t2 := d1, d2
	for head != nil {
		if head.Val < x {
			t1.Next = head
			t1 = t1.Next
		} else {
			t2.Next = head
			t2 = t2.Next
		}
		head = head.Next
	}
	t1.Next = d2.Next
	t2.Next = nil
	return d1.Next
}

  • /**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode} head
 * @param {number} x
 * @return {ListNode}
 */
var partition = function (head, x) {
    const d1 = new ListNode();
    const d2 = new ListNode();
    let t1 = d1,
        t2 = d2;
    while (head) {
        if (head.val < x) {
            t1.next = head;
            t1 = t1.next;
        } else {
            t2.next = head;
            t2 = t2.next;
        }
        head = head.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 head = head;
        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 {
            head = node.next.take();
            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
    }
}

All Problems

All Solutions