LeetCode-496. Next Greater Element I

问题：https://leetcode.com/problems/next-greater-element-i/
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:
All elements in nums1 and nums2 are unique.
The length of both nums1 and nums2 would not exceed 1000.
简单来说就是两个序列，求一个结果序列，第一个序列是第二个序列的子集且元素都unique，第一个序列的每个元素映射到第二个序列的相同的元素，如果第二个序列当前元素的右边出现比此元素大的元素则更新结果序列，否则更新为-1。
分析：使用栈，从后往前遍历nums[i]，每当栈不为空的时候，一直出栈直到遇到比nums[i]大的数字停止。设立一个map

class Solution {public:    vector<int> nextGreaterElement(vector<int>& findNums, vector<int>& nums) {        map<int,int> save;        stack<int> q;        vector<int> res;        for(int i=nums.size()-1;i>=0;i--){            while(!q.empty() && q.top()<nums[i]){                q.pop();            }            save[nums[i]]=q.empty()?-1:q.top();            q.push(nums[i]);        }        for(int i=0;i<findNums.size();i++){            res.push_back(save[findNums[i]]);        }        return res;    }};