496. Next Greater Element I
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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.
#include <iostream>
#include <vector>
using namespace std;
class Solution {
public:
vector<int> nextGreaterElement(vector<int>& findNums, vector<int>& nums) {
vector<int> re;
for(int i=0;i<findNums.size();i++)
{
for(int j=0;j<nums.size();j++)
{
if(findNums[i] == nums[j])
{
if(j != nums.size()-1)
{
int q=j+1;
for(;q<nums.size();q++)
{
if(nums[q] > findNums[i])
{
re.push_back(nums[q]);
break;
}
}
if(q == nums.size())
re.push_back(-1);
}
else if(j == nums.size()-1)
re.push_back(-1);
break;
}
}
}
for(int i=0;i<re.size();i++)
cout << re[i] << endl;
return re;
}
};
int main()
{
Solution s;
vector<int> findNums,nums;
findNums.push_back(2);
findNums.push_back(4);
// findNums.push_back(2);
nums.push_back(1);
nums.push_back(2);
nums.push_back(3);
nums.push_back(4);
s.nextGreaterElement(findNums,nums);
return 0;
}
#include <vector>
using namespace std;
class Solution {
public:
vector<int> nextGreaterElement(vector<int>& findNums, vector<int>& nums) {
vector<int> re;
for(int i=0;i<findNums.size();i++)
{
for(int j=0;j<nums.size();j++)
{
if(findNums[i] == nums[j])
{
if(j != nums.size()-1)
{
int q=j+1;
for(;q<nums.size();q++)
{
if(nums[q] > findNums[i])
{
re.push_back(nums[q]);
break;
}
}
if(q == nums.size())
re.push_back(-1);
}
else if(j == nums.size()-1)
re.push_back(-1);
break;
}
}
}
for(int i=0;i<re.size();i++)
cout << re[i] << endl;
return re;
}
};
int main()
{
Solution s;
vector<int> findNums,nums;
findNums.push_back(2);
findNums.push_back(4);
// findNums.push_back(2);
nums.push_back(1);
nums.push_back(2);
nums.push_back(3);
nums.push_back(4);
s.nextGreaterElement(findNums,nums);
return 0;
}
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- 496. Next Greater Element I
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