<p>You are given the <code>head</code> of a linked list.</p>

<p>Remove every node which has a node with a greater value anywhere to the right side of it.</p>

<p>Return <em>the </em><code>head</code><em> of the modified linked list.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/10/02/drawio.png" style="width: 631px; height: 51px;" />
<pre>
<strong>Input:</strong> head = [5,2,13,3,8]
<strong>Output:</strong> [13,8]
<strong>Explanation:</strong> The nodes that should be removed are 5, 2 and 3.
- Node 13 is to the right of node 5.
- Node 13 is to the right of node 2.
- Node 8 is to the right of node 3.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> head = [1,1,1,1]
<strong>Output:</strong> [1,1,1,1]
<strong>Explanation:</strong> Every node has value 1, so no nodes are removed.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of the nodes in the given list is in the range <code>[1, 10<sup>5</sup>]</code>.</li>
	<li><code>1 &lt;= Node.val &lt;= 10<sup>5</sup></code></li>
</ul>

class Solution {
public:
    ListNode* removeNodes(ListNode* head) {
        stack<ListNode*> st;
        // traverse, push stack, pop until st is empty or 
        ListNode* curr = head;
        while (curr != nullptr)
        {
            while (!st.empty() && st.top()->val < curr->val)
                st.pop();
            st.push(curr);
            curr = curr->next;
        }
        ListNode* new_head, * nex_node = nullptr;
        while (!st.empty())
        {
            ListNode* curr_node = st.top();
            st.pop();
            curr_node->next = nex_node;
            nex_node = curr_node;
            if (st.empty())
                return curr_node;
        }
        return head;
    }
};
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