<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) {
        ListNode* tempA=head;
        ListNode* tempB=head->next;
        vector<int> ans;
        if( head->next==NULL) return {0};
        while(tempA)
        {   tempB=tempA->next;
            while(tempB)
            {
                if(tempB->val > tempA->val)
                {
                     ans.push_back(tempB->val);
                     break;
                }
                tempB=tempB->next;
            }
            if(tempB==NULL) ans.push_back(0);
            tempA=tempA->next;
        }
        return ans;
    }
};
44100