Next Greater Node In Linked List
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1000-1099/1019.Next%20Greater%20Node%20In%20Linked%20List/images/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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1000-1099/1019.Next%20Greater%20Node%20In%20Linked%20List/images/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>
Stack; Array; Linked List; Monotonic Stack
/**
 * 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) {
        vector<int> nums;
        for (; head; head = head->next) {
            nums.push_back(head->val);
        }
        stack<int> stk;
        int n = nums.size();
        vector<int> ans(n);
        for (int i = n - 1; ~i; --i) {
            while (!stk.empty() && stk.top() <= nums[i]) {
                stk.pop();
            }
            if (!stk.empty()) {
                ans[i] = stk.top();
            }
            stk.push(nums[i]);
        }
        return ans;
    }
};