You are given two arrays(without duplicates)nums1andnums2wherenums1’s elements are subset ofnums2. Find all the next greater numbers fornums1's elements in the corresponding places ofnums2.

The Next Greater Number of a numberxinnums1is the first greater number to its right innums2. 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 innums1
   andnums2are unique.
2. The length of bothnums1andnums2would not exceed 1000.

class Solution {
    public int[] nextGreaterElement(int[] nums1, int[] nums2) {
        int n1 = nums1.length;
        int n2 = nums2.length;
        int[] res = new int[n1];
        Arrays.fill(res, -1);
        // List<Integer> list1 = new ArrayList();
        // for(int i = 0; i < n1; i++) list1.
        
        for(int i = 0; i < n1; i++){
            int t = nums1[i];
            int j = 0;
            while(nums2[j] != t) j++;
            j++;
            for(; j < n2; j++){
                if(nums2[j] > t){
                   res[i] = nums2[j];
                    break;
                } 
            }
        }
        return res;
    }
}

什麼狗jb描述，next greater number是說，nums1的element在nums2找到後，往右找有沒有比它更大的，沒有就是-1

class Solution {
    public int[] nextGreaterElement(int[] nums1, int[] nums2) {
        Map<Integer, Integer> map = new HashMap();
        Stack<Integer> stack = new Stack();
        
        for(int i: nums2){
            while(!stack.isEmpty() && stack.peek() < i){
                map.put(stack.pop(), i);
            }
            stack.push(i);
        }
        for(int i = 0; i < nums1.length; i++){
            nums1[i] = map.getOrDefault(nums1[i], -1);
        }
        return nums1;
    }
}

2. O(n)

問的實際上是nums2的next greater number，用stack維持一個decrease的數們，如果當前比peek大，那就pop所有比peek小的，他們的ngn就是當前

用map存放每個數對應的ngn。

https://leetcode.com/problems/next-greater-element-i/discuss/97595/Java-10-lines-linear-time-complexity-O(n)-with-explanation