496.Next Greater Element I

Description:
You are given two arrays (without duplicates) nums1 and nums2 where nums1's elements are subset of nums2. Find all the next greater numbers for nums1's elements in the corresponding places of nums2.

The Next Greater Number of a number x in nums1 is the first greater number to its right in nums2. If it does not exist, output -1 for this number.

Example 1:
Input: nums1 = [4,1,2], nums2 = [1,3,4,2].
Output: [-1,3,-1]
Explanation:
For number 4 in the first array, you cannot find the next greater number for it in the second array, so output -1.
For number 1 in the first array, the next greater number for it in the second array is 3.
For number 2 in the first array, there is no next greater number for it in the second array, so output -1.

Example 2:
Input: nums1 = [2,4], nums2 = [1,2,3,4].
Output: [3,-1]
Explanation:
For number 2 in the first array, the next greater number for it in the second array is 3.
For number 4 in the first array, there is no next greater number for it in the second array, so output -1.

Note:
1. All elements in nums1 and nums2 are unique.
2. The length of both nums1 and nums2 would not exceed 1000.

Train of thought:
1. Find the array elements in nums 1 in nums2.
2. Increase the angle scale and compare the size until the larger element is found in nums 2, or "-1" if it is not found.

Code:

public class Solution {
    public int[] nextGreaterElement(int[] findNums, int[] nums) {
        int[] nextNums=new int[findNums.length];
        int greaterNum=0;

        for(int i=0;i<findNums.length;i++){
            for(int j=0;j<nums.length;j++){
                if(findNums[i]==nums[j]){
                    if(j!=nums.length-1){
                        do{
                            greaterNum=nums[j];
                            if(greaterNum>findNums[i]){
                                nextNums[i]=greaterNum;
                                break;
                            }else if(j==nums.length-1){
                                nextNums[i]=-1;
                            }
                            j++;
                        }while(j<nums.length);
                    }else{
                        nextNums[i]=-1;
                    }
                    break;
                }
            }
        }

        return nextNums;
    }
}

My idea here is to use the corners of the array to do this series of operations.
Stack can also be used to solve this problem:

public class Solution {
    public int[] nextGreaterElement(int[] findNums, int[] nums) {
        Map<Integer, Integer> map = new HashMap<>(); // map from x to next greater element of x
        Stack<Integer> stack = new Stack<>();
        for (int num : nums) {
            while (!stack.isEmpty() && stack.peek() < num)
                map.put(stack.pop(), num);
            stack.push(num);
        }   
        for (int i = 0; i < findNums.length; i++)
            findNums[i] = map.getOrDefault(findNums[i], -1);
        return findNums;
    }
}

This solution greatly reduces the complexity of the algorithm, which is O(n).
Done.