# 86. Partition List
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.
Example:
Input: head = 1->4->3->2->5->2, x = 3
Output: 1->2->2->4->3->5
# Solution
Approach 1: Two pointers.
# Code (Python)
Approach 1:
# Code (C++)
Approach 1:
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* partition(ListNode* head, int x) {
ListNode dummyHead1(0);
ListNode dummyHead2(0);
ListNode *p1 = &dummyHead1;
ListNode *p2 = &dummyHead2;
ListNode *curr = head;
while (curr)
{
if (curr->val < x)
{
p1->next = curr;
p1 = curr;
}
else
{
p2->next = curr;
p2 = curr;
}
curr = curr->next;
}
p1->next = dummyHead2.next;
p2->next = NULL; // to prevent cycle.
return dummyHead1.next;
}
};