Partition List LeetCode Solution | Easy Approach

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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

 Partition List Solutions

Time: O(n)
Space: O(n)

C++

class Solution {
 public:
  ListNode* partition(ListNode* head, int x) {
    ListNode beforeHead(0);
    ListNode afterHead(0);
    ListNode* before = &beforeHead;
    ListNode* after = &afterHead;

    for (; head; head = head->next)
      if (head->val < x) {
        before->next = head;
        before = head;
      } else {
        after->next = head;
        after = head;
      }

    after->next = nullptr;
    before->next = afterHead.next;

    return beforeHead.next;
  };
};

Java

 
class Solution {
  public ListNode partition(ListNode head, int x) {
    ListNode beforeHead = new ListNode(0);
    ListNode afterHead = new ListNode(0);
    ListNode before = beforeHead;
    ListNode after = afterHead;

    for (; head != null; head = head.next)
      if (head.val < x) {
        before.next = head;
        before = head;
      } else {
        after.next = head;
        after = head;
      }

    after.next = null;
    before.next = afterHead.next;

    return beforeHead.next;
  }
}

Python

class Solution:
  def partition(self, head: ListNode, x: int) -> ListNode:
    beforeHead = ListNode(0)
    afterHead = ListNode(0)
    before = beforeHead
    after = afterHead

    while head:
      if head.val < x:
        before.next = head
        before = head
      else:
        after.next = head
        after = head
      head = head.next

    after.next = None
    before.next = afterHead.next

    return beforeHead.next

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