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 lengthk <= m + n from digits of the two. The relative order of the digits from the same array must be preserved. Return an array of thek 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;}};