LeetCode – Partition List (Java)

 

Given 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.

For example, given 1->4->3->2->5->2 and x = 3, return 1->2->2->4->3->5.

Java Solution

public class Solution {
    public ListNode partition(ListNode head, int x) {
        if(head == null) return null;
 
        ListNode fakeHead1 = new ListNode(0);
        ListNode fakeHead2 = new ListNode(0);
        fakeHead1.next = head;
 
        ListNode p = head;
        ListNode prev = fakeHead1;
        ListNode p2 = fakeHead2;
 
        while(p != null){
            if(p.val < x){
                p = p.next;
                prev = prev.next;
            }else{
 
                p2.next = p;
                prev.next = p.next;
 
                p = prev.next;
                p2 = p2.next;
            } 
        }
 
        // close the list
        p2.next = null;
 
        prev.next = fakeHead2.next;
 
        return fakeHead1.next;
    }
}

Related posts:

  1. LeetCode – Reverse Linked List II (Java)
  2. LeetCode – Remove Duplicates from Sorted List
  3. LeetCode – Palindrome Linked List (Java)
  4. LeetCode – Sort List (Java)
Category >> Algorithms  
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  1. Parag Chaudhari on 2015-2-11

    Here is another version,

    public void partitionList(Node start, int num){
    Node p = start;
    Node q = start;
    while(q! = null){
    while(p != null && p.value num)
    q = q.next;
    swapValue(p,q);
    }
    }

  2. Chill on 2015-3-27

    The result gives 1->2->2->4->3->5. It is wrong.
    It has to be 1->2->2->3->4->5.

    Rule: all nodes less than x come before nodes greater than or equal to x.

    Please correct me if I am wrong

  3. Antoine Dahan on 2015-8-13

    Here is a much simpler solution:

  4. Antoine Dahan on 2015-8-13

    Here is a much simpler solution:

    I’m iterating through with two iterators so I can change the current positions previous node’s next (no other way to access it in a singly linked list). Then whenever I encounter a value less than x, I append it in front of the head and set it as the new head. If the value is greater than or equal to x I just go onto the next value.

    After reaching the end, I just return the new head.

    (posted as screenshot because formatting gets messed up in a discuss text post)

  5. Akhil Verghese on 2015-9-21

    Good answer, but if you’re going to swap the p and q values, the q value should be < num. You're skipping passed all the values num), it’ll work (tested and confirmed).

  6. Jeff Guo on 2015-10-18

    This one might be easier to understand. C# code.

    public Node PartitionList(Node head, T number)

    {

    if (head == null || head.Next == null)

    return head;

    var fakeHead1 = new Node(); //smaller ones

    var fakeHead2 = new Node(); //bigger ones

    var p = head;

    var fakeTail1 = fakeHead1;

    var faketail2 = fakeHead2;

    while (p != null)

    {

    if (p.content.CompareTo(number) < 0)

    {

    fakeTail1.Next = p;

    fakeTail1 = fakeTail1.Next;

    p = p.Next;

    fakeTail1.Next = null;

    }

    else

    {

    faketail2.Next = p;

    faketail2 = faketail2.Next;

    p = p.Next;

    faketail2.Next = null;

    }

    }

    fakeTail1.Next = fakeHead2.Next;

    return fakeHead1.Next;

    }

  7. Aditya Vutukuri on 2016-2-25

    The result is correct indeed

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