Question

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.

Solution

  • Method 1: Array O(n^2) 
    class Solution {
    public int[] nextGreaterElements(int[] nums) {
        int[] result = new int[nums.length];
        for(int i = 0; i < nums.length; i++){
            int j = i + 1;
            while(j != i){
                if(j == nums.length){
                    j = 0;
                    if(j == i) break;
                }
                if(nums[j] > nums[i]){
                    result[i] = nums[j];
                    break;
                }
                j++;
            }
            if(j == i) result[i] = -1;
        }
        return result;
    }
}