# 503. Next Greater Element II

Given a circular integer array `nums` (i.e., the next element of `nums[nums.length - 1]` is `nums[0]`), return *the **next greater number** for every element in* `nums`.

The **next greater number** of a number `x` 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't exist, return `-1` for this number.

**Example 1:**

<pre><code><strong>Input: nums = [1,2,1]
</strong><strong>Output: [2,-1,2]
</strong>Explanation: The first 1's next greater number is 2; 
The number 2 can't find next greater number. 
The second 1's next greater number needs to search circularly, which is also 2.
</code></pre>

**Example 2:**

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

**Constraints:**

* `1 <= nums.length <= 104`
* `-109 <= nums[i] <= 109`

***My Solutions:***

和496类似，但是需要遍历数组两次。遇到比栈顶小的数字就放入栈中；遇到比栈顶大的数字就说明此时的数字是栈顶数字的下一个更大的数，就将其放在结果数组的对应位置上，栈顶的元素出栈。继续比较新的栈顶的数，如果还是大，那么继续记录，出栈，直到栈顶的数比新数要小，那么就可以将新数入栈了。因为要将找到的更大的数放在对应位置上，所以栈中记录的应该是元素位置，而不是具体的数字。因为 res 的长度必须是 n，超过 n 的部分我们只是为了给之前栈中的数字找较大值，所以不能压入栈。

```java
class Solution {
    public int[] nextGreaterElements(int[] nums) {
        int len = nums.length;
        int[] res = new int[len];
        Arrays.fill(res, -1);
        Stack<Integer> stack = new Stack<>();
        for (int i = 0; i < len * 2; i++) {
            int index = i % len;
            while (!stack.isEmpty() && nums[stack.peek()] < nums[index]) {
                res[stack.peek()] = nums[index];
                stack.pop();
            }
            if (i < len) stack.push(index);
        }
        return res;
    }
}
```
