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# Partition List LeetCode Solution

## Problem – Partition List LeetCode Solution

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 LeetCode Solution in Python

``````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
``````

## Partition List LeetCode Solution in Java

``````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;
}
}
``````

## Partition List LeetCode Solution in C++

``````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;
}
};
``````
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