leetcode.321. Create Maximum Number

Given two arrays of length

m

and

n

with digits

0-9

representing two numbers. Create the maximum number of length

k <= m + n

from digits of the two. The relative order of the digits from the same array must be preserved. Return an array of the

k

digits. You should try to optimize your time and space complexity.
Example 1:

nums1 =

[3, 4, 6, 5]

nums2 =

[9, 1, 2, 5, 8, 3]

k =

5

return

[9, 8, 6, 5, 3]

Example 2:

nums1 =

[6, 7]

nums2 =

[6, 0, 4]

k =

5

return

[6, 7, 6, 0, 4]

Example 3:

nums1 =

[3, 9]

nums2 =

[8, 9]

k =

3

return

[9, 8, 9]

class Solution {
public:
vector<int> maxNumber(vector<int>& nums1, vector<int>& nums2, int k) {
int size1 = nums1.size();
int size2 = nums2.size();
vector<int> res(k);
vector<int> tmp(k);
for (int i = 0; i <= k; i++)
{
if (k - i > size2)  continue;
if (i > size1)        break;
vector<int> num1;
vector<int> num2;
num1 = getnum(nums1, i);
num2 = getnum(nums2, k - i);
tmp = mergenum(num1, num2);
res = cmp(res, tmp);
}
return res;
}
vector<int> getnum(vector<int>& num, int n)
{
vector<int> vec;
int drop = num.size()-n;
for(int i =0; i< num.size(); i++)
{
while(drop >0 && vec.size() && num[i] > vec.back())
{
drop--;
vec.pop_back();
}
vec.push_back(num[i]);
}
vec.resize(n);
return vec;

}
vector<int> mergenum(vector<int> & num1, vector<int> & num2)
{
vector<int> vec;
while(num1.size() + num2.size() > 0)
{
vector<int>& tem = num1>num2?num1:num2;
vec.push_back(tem[0]);
tem.erase(tem.begin());
}
return vec;

}
vector<int> cmp(vector<int> & num1, vector<int> & num2)
{
int i = 0;
int j = 0;
while (i < num1.size() && j < num2.size())
{
if (num1[i]>num2[j])
return num1;
if (num1[i] < num2[j])
return num2;

i++;
j++;
}
return num1;
}

};