<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:
ListNode* reverseList(ListNode* head) {
    if(head == NULL  || head->next == NULL) return head;
    ListNode* newHead = reverseList(head->next);
    head->next->next = head;
    head->next = NULL;
    return newHead;
}
 vector<int> nextLargerNodes(ListNode* head) {
        stack<int> st;
        int size =0;
        ListNode* temp = head;
        while(temp){
            temp = temp->next;
            size++;
        } 
         head= reverseList(head);

        temp = head;
        vector<int> v(size,0);
        st.push(temp->val);
        v[0] = 0;
        temp = temp->next;
        for(int i=1;i<size;i++){
            while(st.size() != 0 && st.top() <= temp->val ) {
                st.pop();
        }
        if(st.size()== 0) v[i] = 0;
        else v[i] = st.top();
        st.push(temp->val);
        temp = temp->next;
        
    }
    reverse(v.begin(),v.end());
     return v;
    }
   
};
41000