<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:
    int Length(ListNode*head){
        int len = 0;
        while(head){
        len++;
        head = head->next;
        }
        return len;
    }
    ListNode* ReverseLL(ListNode* head){  // REVERSE LL
        if(head == NULL || head->next == NULL)
            return head;
        ListNode* newhead = ReverseLL(head->next);
        head->next->next = head;
        head->next = NULL;
        return newhead;
    }
    vector<int> nextLargerNodes(ListNode* head) { // SINCE WE  reversed ll find prev greater Node
       stack<int>st;
       int len = Length(head);
       vector<int>v(len , 0);
       head = ReverseLL(head);
        int i = 0;
       while(head){
            while(!st.empty() && head->val >= st.top())
                st.pop();
            if(!st.empty())  v[i] = st.top();
            st.push(head->val);
            head = head->next;   
            i++;
       }
       reverse(v.begin() , v.end()); // reverse the vector
       return v;
    }
};

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