Next Greater Element II
<p>Given a circular integer array <code>nums</code> (i.e., the next element of <code>nums[nums.length - 1]</code> is <code>nums[0]</code>), return <em>the <strong>next greater number</strong> for every element in</em> <code>nums</code>.</p>

<p>The <strong>next greater number</strong> of a number <code>x</code> is the first greater number to its traversing-order next in the array, which means you could search circularly to find its next greater number. If it doesn&#39;t exist, return <code>-1</code> for this number.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,1]
<strong>Output:</strong> [2,-1,2]
Explanation: The first 1&#39;s next greater number is 2; 
The number 2 can&#39;t find next greater number. 
The second 1&#39;s next greater number needs to search circularly, which is also 2.
</pre>

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

<pre>
<strong>Input:</strong> nums = [1,2,3,4,3]
<strong>Output:</strong> [2,3,4,-1,4]
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>4</sup></code></li>
	<li><code>-10<sup>9</sup> &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
</ul>
Stack; Array; Monotonic Stack
class Solution {
    public int[] nextGreaterElements(int[] nums) {
        int n = nums.length;
        int[] ans = new int[n];
        Arrays.fill(ans, -1);
        Deque<Integer> stk = new ArrayDeque<>();
        for (int i = n * 2 - 1; i >= 0; --i) {
            int j = i % n;
            while (!stk.isEmpty() && stk.peek() <= nums[j]) {
                stk.pop();
            }
            if (!stk.isEmpty()) {
                ans[j] = stk.peek();
            }
            stk.push(nums[j]);
        }
        return ans;
    }
}