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
andnums2
are unique. - The length of both
nums1
andnums2
would not exceed 1000.
class Solution {
public:
vector<int> nextGreaterElement(vector<int>& findNums, vector<int>& nums) {
stack <int>s;
unordered_map<int,int>m;
vector <int>res;
for(int n:nums){
while(!s.empty()&& n>s.top()){
m[s.top()]=n;
s.pop();
}
s.push(n);
}
for(int i: findNums){
if(m.count(i)){
res.push_back(m[i]);
}
else
res.push_back(-1);
}
return res;
}
};