Given a circular array (the next element of the last element is the first element of the array), print the Next Greater Number for every element. 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, output -1 for this number.

Example 1:

Input: [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.

我們可以使用棧來進行優化上面的算法，我們遍歷兩倍的數組，然後還是坐標i對n取余，取出數字，如果此時棧不為空，且棧頂元素小於當前數字，說明當前數字就是棧頂元素的右邊第一個較大數，那麽建立二者的映射，並且去除當前棧頂元素，最後如果i小於n，則把i壓入棧。因為res的長度必須是n，超過n的部分我們只是為了給之前棧中的數字找較大值

沒有用hashMap 存-1的值, 用的是Arrays.fill(nums, -1)

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

503. Next Greater Element II