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