Title
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.
哎，这个题目还是挺难理解的，就是给出一组数据，求每个元素右边的大于它的第一个数字，用map得到每个数字的对应数字，使用栈保存一组数字，当遍历数组时，把当前数字作为结果的全部用map存储，并pop出去
Solution
//cpp:
class Solution {
public:
vector<int> nextGreaterElement(vector<int>& findNums, vector<int>& nums) {
unordered_map<int, int> map1;
stack<int> s;
for (int num : nums) {
map1[num] = -1;
while (!s.empty() && s.top() < num) {
map1[s.top()] = num;
s.pop();
}
s.push(num);
}
vector<int> v;
for (auto it : findNums) {
v.push_back(map1[it]);
//ou(it);
}
return v;
}
};