Problem Statement

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

Problem Link

Partition List - LeetCode

Code

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* partition(ListNode* head, int x) {
        
        if(!head)
            return head;
        
        ListNode node1(0);
        ListNode node2(0);
        
        ListNode *p1 = &node1, *p2 = &node2;
        
        while(head)
        {
            if(head->val<x)
            {
                p1->next = head;
                p1 = p1->next;
            }
            else
            {
                p2->next = head;
                p2 = p2->next;
            }
            
            head = head->next;
        }
        
        p1->next = node2.next;
        p2->next = NULL;
        
        return node1.next;
        
        
        
    }
};