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Partition List

Updated: Apr 14, 2021

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:

/**
 * 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 head1 = new ListNode(-1);
        ListNode head2 = new ListNode(-1);
        ListNode small = head1;
        ListNode large = head2;
        
        ListNode current = head;
        while(current!=null)
        {
            if(current.val<x)
            {
                small.next=current;
                small=current;
            }
            else
            {
                large.next = current;
                large=current;
            }
            current = current.next;
        }
        large.next=null;
        small.next = head2.next;
        return head1.next;
    }
}

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