496. Next Greater Element I ----------E ~~~~

 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.

A:

哎，第二次，竟然没有想起来怎么做。

虽然也想到了用increasing stack， 可是还是不够理想

class Solution {
public:
    vector<int> nextGreaterElement(vector<int>& nums1, vector<int>& nums2) {
        // keep an increaseing stack of nums2, from back to front
        stack<int> minStack; // top is the smallest
        int n1 = nums1.size(), n2 = nums2.size();
        unordered_map<int, int> M; // value to its next grater
        for(int i = n2-1;i>=0;i--){
            while(not minStack.empty() and minStack.top() <= nums2[i]){
                minStack.pop();
            }
            M[nums2[i]] = minStack.empty()?-1:minStack.top();            
            minStack.push(nums2[i]);
        }
        vector<int> res;
        for(int i =0;i<n1;i++)
            res.push_back(M[nums1[i]]);
        return res;
    }
};