# Leetcode Merge Sorted Array problem solution

In this Leetcode Merge Sorted Array problem solution, You are given two integer arrays nums1 and nums2, sorted in non-decreasing order, and two integers m and n, representing the number of elements in nums1 and nums2 respectively. Merge nums1 and nums2 into a single array sorted in non-decreasing order.

The function should not return the final sorted array, but instead, be stored inside the array nums1. To accommodate this, nums1 has a length of m + n, where the first m elements denote the elements that should be merged, and the last n elements are set to 0 and should be ignored. nums2 has a length of n.

## Problem solution in Python.

```def merge(self, nums1, m, nums2, n):
i=0
for x in range(len(nums1)):
if i>=n:
break
if nums1[x]==0:
nums1[x]=nums2[i]
i+=1
nums1.sort()
```

## Problem solution in Java.

```public void merge(int A[], int m, int B[], int n) {
int insertIndex = m+n-1;
int indexA = m-1,indexB = n-1;

while(indexB>=0){
if(indexA<0){
A[insertIndex--] = B[indexB--];
}else{
if(B[indexB]>=A[indexA]){
A[insertIndex--] = B[indexB--];
}else{
A[insertIndex--] = A[indexA--];
}
}
}
}
```

## Problem solution in C++.

```class Solution {
public:
void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {
int i=0, j=0;
vector<int> nums;
while(1)
{
if(i==m)
{
while(j<n)
{
nums.push_back(nums2[j]);
j++;
}
break;
}

else if(j==n)
{
while(i<m)
{
nums.push_back(nums1[i]);
i++;
}
break;
}

else
{
if(nums1[i] < nums2[j])
{
nums.push_back(nums1[i]);
i++;
}
else
{
nums.push_back(nums2[j]);
j++;
}
}
}

for(i=0; i<m+n; i++)
nums1[i] = nums[i];
}
};
```

## Problem solution in C.

```int compare (int *a, int *b){

return *a > *b;
}

void merge(int* nums1, int nums1Size, int m, int* nums2, int nums2Size, int n){

int i;

for(i = 0; i < n; i++){

nums1[m + i] = nums2[i];
}

qsort(nums1, nums1Size, sizeof(int), compare);

}
```