Solution: Partition List

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Leetcode Problem #86 (Medium): Partition List

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Description:

(Jump to: Solution Idea || Code: JavaScript | Python | Java | C++)

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.

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Examples:

Example 1:
Input:	head = [1,4,3,2,5,2], x = 3
Output:	[1,2,2,4,3,5]
Visual:

Example 2:
Input:	head = [2,1], x = 2
Output:	[1,2]

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Constraints:

• The number of nodes in the list is in the range [0, 200].
• -100 <= Node.val <= 100
• -200 <= x <= 200

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Idea:

(Jump to: Problem Description || Code: JavaScript | Python | Java | C++)

The easiest thing to do here is to create separate linked lists for the front and back portions of list we want to return. In order to accomplish that, we should first create some dummy heads (fdum, bdum), then create pointers for the current nodes each of the front, back, and main lists (front, back, curr).

Then we can simply iterate through the main list and stitch together each node to either front or back, depending on the node's value.

Once we reach the end, we just need to stitch together the two sub-lists, making sure to cap off the end of back, and then return our new list, minus the dummy head.

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Implementation:

There are only minor differences between the code of all four languages.

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Javascript Code:

(Jump to: Problem Description || Solution Idea)

var partition = function(head, x) {
    let fdum = new ListNode(0), bdum = new ListNode(0),
        front = fdum, back = bdum, curr = head
    while (curr) {
        if (curr.val < x)front.next = curr, front = curr
        else back.next = curr, back = curr
        curr = curr.next
    }
    front.next = bdum.next, back.next = null
    return fdum.next
};

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Python Code:

(Jump to: Problem Description || Solution Idea)

class Solution:
    def partition(self, head: ListNode, x: int) -> ListNode:
        fdum, bdum = ListNode(0), ListNode(0)
        front, back, curr = fdum, bdum, head
        while curr:
            if curr.val < x:
                front.next = curr
                front = curr
            else:
                back.next = curr
                back = curr
            curr = curr.next
        front.next, back.next = bdum.next, None
        return fdum.next

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Java Code:

(Jump to: Problem Description || Solution Idea)

class Solution {
    public ListNode partition(ListNode head, int x) {
        ListNode fdum = new ListNode(0), bdum = new ListNode(0),
                 front = fdum, back = bdum, curr = head;
        while (curr != null) {
            if (curr.val < x) {
                front.next = curr;
                front = curr;
            } else {
                back.next = curr;
                back = curr;
            }
            curr = curr.next;
        }
        front.next = bdum.next;
        back.next = null;
        return fdum.next;
    }
}

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C++ Code:

(Jump to: Problem Description || Solution Idea)

class Solution {
public:
    ListNode* partition(ListNode* head, int x) {
        ListNode *fdum = new ListNode(0), *bdum = new ListNode(0),
                 *front = fdum, *back = bdum, *curr = head;
        while (curr) {
            if (curr->val < x) front->next = curr, front = curr;
            else back->next = curr, back = curr;
            curr = curr->next;
        }
        front->next = bdum->next, back->next = nullptr;
        return fdum->next;
    }
};

Comments

[Aaron Reese]

In js at least I would use array.filter into two new arrays and then use the spread operator to merge the arrays. 4 lines of clean code, or 1 line of horrible code if you are a smart-arse Dev 😁

[seanpgallivan]

Unfortunately, the input is a linked list, not an array. It's shown as an array in the example just to make it easier to read.