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.

數組1是數組2的子數組，找出數組1中元素在數組2中同位置之後第一個大於該元素的元素。

只要找到數組2中每一位元素後面比它大的第一個元素即可。因為數組1是數組2的子數組，並且是按索引比較。

利用stack和map即可。

map用來存儲每個元素後面第一個大於其值的元素，stack進行遍歷，求出每一個大於元素。

class Solution {
public:
    vector<int> nextGreaterElement(vector<int>& findNums, vector<int>& nums) {
        vector<int> res;
        stack<int> s;
        unordered_map<int, int> m;
        for (int& num : nums)
        {
            while (!s.empty() && s.top() < num)
            {
                m[s.top()] = num;
                s.pop();
            }
            s.push(num);
        }
        int tmp = 0;
        for (int& n : findNums)
        {
            tmp = m.count(n) ? m[n] : -1;
            res.push_back(tmp);
        }
        return res;
    }
};

[LeetCode] Next Greater Element I