503. Next Greater Element II 首尾可循环

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;
    }