LeetCode

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.

Note: The length of given array won't exceed 10000.

class Solution {
    public int[] nextGreaterElements(int[] nums) {
        if (nums == null || nums.length <= 0)
            return new int[0];
        int[] ret = new int[nums.length];
        for (int i=0; i<ret.length; i++)
            ret[i] = -1;
        for (int i=0; i<nums.length; i++) {
            int flag = 0;
            for (int j=i+1; j<nums.length; j++) {
                if (nums[j] > nums[i]) {
                    ret[i] = nums[j];
                    flag = 1;
                    break;
                }
            }
            if (flag == 0) {
                for (int k=0; k<i; k++) {
                    if (nums[k] > nums[i]) {
                        ret[i] = nums[k];
                        break;
                    }
                }
            }
        }
        return ret;
    }
}