# 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;
    }
};