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:
nums1
and nums2
are unique.nums1
and nums2
would not exceed 1000.class Solution {
public:
vector
std::unordered_map
vector
for (int i = 0; i < nums.size(); i++)
{
int a = nums[i];
m[a] = i;
}
for (int i = 0; i < findNums.size(); i++)
{
int a = findNums[i];
if (m.find(a) != m.end()) {
int j = m[a]+1;
while (j < nums.size())
{
if (nums[j] > a) {
vi.push_back(nums[j]);
break;
}
j++;
}
if (j >= nums.size()) {
vi.push_back(-1);
}
}
}
return vi;
}
};