<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>

class Solution {
public:
    vector<int> nextLargerNodes(ListNode* head) {
        int temp_max = 0;

        stack<int> st1;

        while(head != NULL) {
            st1.push(head->val);
            head = head->next;
        }

        vector<int> ret;
        stack<int> st2;

        while(!st1.empty()) {
            int cur = st1.top();
            st1.pop();

            if (st2.empty()) {
                ret.push_back(0);
                st2.push(cur);
            } else {
                while (!st2.empty()) {
                    if (cur >= st2.top())
                        st2.pop();
                    else
                        break;
                }

                if (!st2.empty())
                    ret.push_back(st2.top());
                else
                    ret.push_back(0);
                
                st2.push(cur);
            }
        }

        reverse(ret.begin(), ret.end());

        return ret;
    }
};
45300