Brute Force

O(n^2)

• one way to solve circular array is to extend the length of array to twice length

public class Solution {
    public int[] nextGreaterElements(int[] nums) {

        int[] tmp = new int[nums.length * 2];

        int index = 0;
        for (int i = 0; i < 2; ++i) {
            for (int j = 0; j < nums.length; ++j) {
                tmp[index++] = nums[j];
            }
        }

        int[] result = new int[nums.length];

        for (int i = 0; i < nums.length; ++i) {
            int j = i + 1;

            while (j < tmp.length && tmp[j] <= nums[i]) {
                j++;
            }
            result[i] = j == tmp.length ? -1 : tmp[j];
        }

        return result;
    }
}

get rid of tmp array

public int[] nextGreaterElements2(int[] nums) {
    int  n = nums.length;
    int[] res = new int[n];
    for (int i = 0; i < n; i++) {
        res[i] = -1;
        for (int k = i+1; k < i + n; k++) {
            if (nums[k%n] > nums[i]){
                res[i] = nums[k%n];
                break;
            }
        }
    }
    return res;
}

Stack

O(n)

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