503. Next Greater Element II

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.


// solution 1 : a working solution 
typical way to solve circular array problems is to extend the original array to twice length, 2nd half has the same element as first half.

class Solution {
    public int[] nextGreaterElements(int[] nums) {
        //  6 4 5 1 2 1 , 6 4 5 1 2 1
        // -1 5 5 2 6 6 
        // add anothe array of the same value 
        // traverse the original array, and look for the next element in the new combined array
        // if not found until found itself again, return -1, break 
        
        int[] newArray = new int[nums.length * 2];
        for(int i = 0; i < nums.length; i++){
            newArray[i] = nums[i];
        }
        for(int i = nums.length; i < newArray.length; i++){
            newArray[i] = nums[i - nums.length];
        }
        
        int[] res = new int[nums.length];
        for(int i = 0; i < nums.length; i++){
            res[i] = helper(newArray, nums[i], i);
        }
        return res;
    }
    private int helper(int[] newArray, int num, int index){
        for(int i = index + 1; i < newArray.length; i++){
            if(newArray[i] > num){
                return newArray[i];
            }
        }
        return -1;
        
    }
}


// use stack 

class Solution {
    public int[] nextGreaterElements(int[] nums) {
        int[] res = new int[nums.length];
        Stack<Integer> stack = new Stack<>();
        // like in solution1 , extend the num by length * 2 
        // push the nums into the stack from 1 to 6 
        // original array : 6 4 5 1 2 1 
        // same process as we did in solution when using a stack 
        for(int i = nums.length - 1; i >= 0; i--){
            stack.push(nums[i]);
        }
        for(int i = nums.length - 1; i >= 0; i--){
            while(!stack.isEmpty() && stack.peek() <= nums[i]){ // = also included, same element, also returns -1 
                stack.pop();
            }
            if(stack.isEmpty()){
                res[i] = -1;
            }else{
                res[i] = stack.peek();
            }
            stack.push(nums[i]);
        }
        return res;
    }
}