<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> vals;
        vector<int> result; // Store the final results
        
        // First, convert the linked list into an array of values
        ListNode* cur = head;
        while (cur) {
            vals.push_back(cur->val);
            result.push_back(0); // Initialize result array with 0
            cur = cur->next;
        }
        
        // Use a stack to store indices whose next larger element is not found yet
        stack<int> st;
        
        // Traverse through the vals array
        for (int i = 0; i < vals.size(); ++i) {
            // While the stack is not empty and the current value is larger than the value at the top of the stack
            while (!st.empty() && vals[i] > vals[st.top()]) {
                result[st.top()] = vals[i];  // Update the result at that index
                st.pop(); // Remove the index from the stack
            }
            st.push(i); // Push the current index onto the stack
        }
        
        return result;
    }
};

auto init = []() 
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    return 0;
}();

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