[LeetCode] 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:

1. All elements in nums1 and nums2 are unique.
2. The length of both nums1 and nums2 would not exceed 1000.

数组1是数组2的子数组，找出数组1中元素在数组2中同位置之后第一个大于该元素的元素。

只要找到数组2中每一位元素后面比它大的第一个元素即可。因为数组1是数组2的子数组，并且是按索引比较。

利用stack和map即可。

map用来存储每个元素后面第一个大于其值的元素，stack进行遍历，求出每一个大于元素。

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;
    }
};