[LeetCode] 496. 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.

/* * 依次将nums中的元素添加入stack中，并维持一个递减数列。 * 当欲添加到stack中的元素，比栈顶的几个元素大的时候， * 那么我们就得到了栈顶几个元素的Next Greater Element */class Solution {public:    vector<int> nextGreaterElement(vector<int>& findNums, vector<int>& nums) {        stack<int> stk;        unordered_map<int, int> NxtGrt;        for (auto num : nums) {            while (!stk.empty() && stk.top() < num) {                NxtGrt[stk.top()] = num;                stk.pop();            }            stk.push(num);        }        for (auto& findNum : findNums)            findNum = NxtGrt.find(findNum) == NxtGrt.end() ? -1 : NxtGrt[findNum];        return findNums;    }};