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.

一刷
題解：與496不一樣，數(shù)組變成了首位相連的數(shù)組。于是for-loop的范圍變成了2*len. 方法與496一樣。

public class Solution {
    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;
    }
}