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# 88. Merge Sorted Array

## LeetCode [88. Merge Sorted Array](https://github.com/txfs19260817/LeetCode-Go/blob/master/solutions/title/README.md)

### Description

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

The number of elements initialized in `nums1` and `nums2` are `m` and `n` respectively. You may assume that `nums1` has a size equal to `m + n` such that it has enough space to hold additional elements from `nums2`.

**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]
```

**Example 2:**

```
Input: nums1 = [1], m = 1, nums2 = [], n = 0
Output: [1]
```

**Constraints:**

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

### Tags

Array, Two Pointers

### Solution

Use two pointers `i, j` start from `m - 1` at `nums1` and `n - 1` at `nums2` respectively, and another pointer `k` is assigned to the last position of `nums1`. Compare nums1\[i] and nums2\[j] until one of both pointers is out of index. We move the larger one to the slot which `k` points to, and decrease `k` and the pointer of the larger one by 1 respectively. After this loop, we still need to check if there are remaining numbers in `nums2` (we do not need to check `nums1` because elements of `nums1` are already in their place). We repeatedly assign `nums2[j]` to `nums1[k]` until `j` is out of index.

### Complexity

* Time complexity: $$O(n)$$
* Space complexity: $$O(1)$$

### Code

```go
func merge(nums1 []int, m int, nums2 []int, n int) {
	i, j, k := m-1, n-1, m+n-1
	for ; i >= 0 && j >= 0; k-- {
		if nums1[i] > nums2[j] {
			nums1[k] = nums1[i]
			i--
		} else {
			nums1[k] = nums2[j]
			j--
		}
	}
	for ; j >= 0; k-- {
		nums1[k] = nums2[j]
		j--
	}
}
```