# Description

Given two sorted integer arrays nums1 and nums2, merge nums2 into nums1 as one sorted array.

Note:
You may assume that nums1 has enough space (size that is greater or equal to m + n) to hold additional elements from nums2. The number of elements initialized in nums1 and nums2 are m and n respectively.

The original problem is here.

The original code is here.

# My Solution

I solve this problem in C++, as below:

``````#include<iostream>
#include<math.h>
#include<vector>
#include<stdlib.h>
using namespace std;

class Solution {
public:
void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {
vector<int> nums3;
vector<int>::iterator  iter1, iter2,iter3;
int count=0,count1=0, count2=0;
int temp = 0;
iter1=nums1.begin();
for(count1=0;count1<m;count1++){
nums3.push_back(*iter1);
iter1++;
}
iter1=nums1.begin();
iter2=nums2.begin();
iter3=nums3.begin();
count1=0;
for(;iter2<nums2.end()&&iter3<nums3.end();){
cout<<*iter3<<" "<<*iter2<<endl;
if(*iter3 <= *iter2){
*iter1 = *iter3;
iter1 ++;
iter3 ++;
}
else{
*iter1 = *iter2;
iter1 ++;
iter2 ++;
}
}
while(iter2<nums2.end()){
*iter1 = *iter2;
iter1++;
iter2++;
}
while(iter3<nums3.end()){
*iter1 = *iter3;
iter1++;
iter3++;
}
}
};
// The code under blow is used for test
int main()
{
vector<int> nums1;
vector<int> nums2;
vector<int>::iterator iter;
Solution s;
nums1.push_back(2);
nums1.push_back(0);
nums2.push_back(1);
s.merge(nums1,1,nums2,1);
for(iter=nums1.begin();iter<nums1.end();++iter){
cout<<*iter<<endl;
}
system("pause");
return 0;
}
``````

# Note

To simplify the problem, I copy the useful values from nums1 to nums3. Then, compare the nums2 and nums3, and using “merge sort” algorithm to get the sorted array nums1.