# 0088. Merge Sorted Array

https://leetcode.com/problems/merge-sorted-array

## Description

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 final sorted array should not be returned by the function, 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`.

**Example 1:**

```
**Input:** nums1 = [1,2,3,0,0,0], m = 3, nums2 = [2,5,6], n = 3
**Output:** [1,2,2,3,5,6]
**Explanation:** The arrays we are merging are [1,2,3] and [2,5,6].
The result of the merge is [1,2,2,3,5,6] with the underlined elements coming from nums1.
```

**Example 2:**

```
**Input:** nums1 = [1], m = 1, nums2 = [], n = 0
**Output:** [1]
**Explanation:** The arrays we are merging are [1] and [].
The result of the merge is [1].
```

**Example 3:**

```
**Input:** nums1 = [0], m = 0, nums2 = [1], n = 1
**Output:** [1]
**Explanation:** The arrays we are merging are [] and [1].
The result of the merge is [1].
Note that because m = 0, there are no elements in nums1. The 0 is only there to ensure the merge result can fit in nums1.
```

**Constraints:**

* `nums1.length == m + n`
* `nums2.length == n`
* `0 <= m, n <= 200`
* `1 <= m + n <= 200`
* `-109 <= nums1[i], nums2[j] <= 109`

**Follow up:** Can you come up with an algorithm that runs in `O(m + n)` time?

## ac

```java
class Solution {
public:
    void merge(int A[], int m, int B[], int n) {

        int a=m-1;
        int b=n-1;
        int i=m+n-1;    // calculate the index of the last element of the merged array

        // go from the back by A and B and compare and put to the A element which is larger
        while(a>=0 && b>=0){
            if(A[a]>B[b])   A[i--]=A[a--];
            else            A[i--]=B[b--];
        }

        // if B is longer than A just copy the rest of B to A location, otherwise no need to do anything
        while(b>=0)         A[i--]=B[b--];
    }
};
```
