<p>You are given the <code>head</code> of a linked list with <code>n</code> nodes.</p>

<p>For each node in the list, find the value of the <strong>next greater node</strong>. That is, for each node, find the value of the first node that is next to it and has a <strong>strictly larger</strong> value than it.</p>

<p>Return an integer array <code>answer</code> where <code>answer[i]</code> is the value of the next greater node of the <code>i<sup>th</sup></code> node (<strong>1-indexed</strong>). If the <code>i<sup>th</sup></code> node does not have a next greater node, set <code>answer[i] = 0</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/08/05/linkedlistnext1.jpg" style="width: 304px; height: 133px;" />
<pre>
<strong>Input:</strong> head = [2,1,5]
<strong>Output:</strong> [5,5,0]
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/08/05/linkedlistnext2.jpg" style="width: 500px; height: 113px;" />
<pre>
<strong>Input:</strong> head = [2,7,4,3,5]
<strong>Output:</strong> [7,0,5,5,0]
</pre>

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

<ul>
	<li>The number of nodes in the list is <code>n</code>.</li>
	<li><code>1 &lt;= n &lt;= 10<sup>4</sup></code></li>
	<li><code>1 &lt;= Node.val &lt;= 10<sup>9</sup></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:
    // vector<int> nextLargerNodes(ListNode* head) {
    //     vector<int> v;
    //     ListNode* temp = head;
    //     while(temp!=NULL){
    //         v.push_back(temp->val);
    //         temp = temp->next;
    //     }
    //     int n = v.size();
    //     vector<int> ans(n);
    //     ans[n-1] = 0;
    //     int mx = 0;
    //     if(v[n-1]>v[n-2]) mx = v[n-1];
    //     ans[n-2] = mx;
    //     for(int i=n-3; i>=0; i--){
    //         if(v[i]>mx){
    //             mx = v[i];
    //             ans[i] = 0;
    //         }
    //         else ans[i] = mx;
    //     }
    //     return ans;
    // }
    vector<int> nextLargerNodes(ListNode* head) {
        // Step 1: Convert the linked list into an array (vector)
        vector<int> v;
        ListNode* temp = head;
        while (temp != nullptr) {
            v.push_back(temp->val);
            temp = temp->next;
        }

        int n = v.size();
        vector<int> ans(n, 0);  // Initialize the result array with 0s

        // Step 2: Brute force approach to find the next greater element
        for (int i = 0; i < n; ++i) {
            for (int j = i + 1; j < n; ++j) {
                if (v[j] > v[i]) {
                    ans[i] = v[j];
                    break;  // Stop once the next greater element is found
                }
            }
        }

        return ans;
    }

};
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