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.
idea:
区别是需要遍历两次array，i对n取余得到index，stack里存的是index。没了。
Stack<Integer> s = new Stack<>();
for (int i = 0; i < 2 * n; i++) {
int temp = nums[i % n];
while (!s.isEmpty() && nums[s.peek()] < temp) {
re[s.pop()] = temp;
}
if (i < n)s.push(i);
}
