# 0503. Next Greater Element II

<https://leetcode.com/problems/next-greater-element-ii>

## Description

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:**

```
**Input:** nums = [1,2,1]
**Output:** [2,-1,2]
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.
```

**Example 2:**

```
**Input:** nums = [1,2,3,4,3]
**Output:** [2,3,4,-1,4]
```

**Constraints:**

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

## ac

```java
class Solution {
    public int[] nextGreaterElements(int[] nums) {
        // edge cases
        if (nums == null || nums.length == 0) return new int[0];

        int n = nums.length;
        int[] res = new int[n];
        Arrays.fill(res, -1);
        Stack<Integer> stack = new Stack<>();
        for (int i = 0; i < n * 2; i++) {
            while (!stack.isEmpty() && nums[i%n] > nums[stack.peek()]) {
                int tmp = stack.pop();
                res[tmp] = nums[i%n];
            }
            stack.push(i%n);
        }

        return res;
    }
}
```


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