<p>Given the <code>head</code> of a linked list and a value <code>x</code>, partition it such that all nodes <strong>less than</strong> <code>x</code> come before nodes <strong>greater than or equal</strong> to <code>x</code>.</p>

<p>You should <strong>preserve</strong> the original relative order of the nodes in each of the two partitions.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/01/04/partition.jpg" style="width: 662px; height: 222px;" />
<pre>
<strong>Input:</strong> head = [1,4,3,2,5,2], x = 3
<strong>Output:</strong> [1,2,2,4,3,5]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> head = [2,1], x = 2
<strong>Output:</strong> [1,2]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the list is in the range <code>[0, 200]</code>.</li>
	<li><code>-100 &lt;= Node.val &lt;= 100</code></li>
	<li><code>-200 &lt;= x &lt;= 200</code></li>
</ul>

/**
 * 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) {
        vector<int> arr1;

        vector<int> arr2;

        ListNode* temp = head;

        while(temp != nullptr){
            if(temp->val <x ){
                 arr1.push_back(temp->val);
            }
            else{
                arr2.push_back(temp->val);
            }
            temp = temp->next;
        }

        int i = 0 ; 
        int j = 0 ; 
        int n = arr1.size();
        int m = arr2.size();
        temp = head;
       // cout<<i<<" "<< j<< " "<<n <<" "<< m<<endl;
        // for(auto t: arr1){
        //     cout<<t<<endl;
        // }

        // for(auto t:arr2){
        //     cout<<t<<endl;
        // }
        while(i < n || j< m){

            if( i != n){
               temp->val = arr1[i];
               i++;
            }
            else{
                temp->val = arr2[j];
                j++;
            }
            temp= temp->next;
        }
        return head;
    }
};
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