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# 88. Merge Sorted Array

Given two sorted integer arrays *nums1 \_and \_nums2*, merge \_nums2 \_into \_nums1 \_as one sorted array.

**Note:**

* The number of elements initialized in \_nums1 \_and \_nums2 \_are \_m \_and \_n \_respectively.
* You may assume that *nums1 \_has enough space (size that is greater or equal to \_m* +*n*) to hold additional elements from *nums2*.

**Example:**

```
Input:

nums1 = [1,2,3,0,0,0], m = 3
nums2 = [2,5,6],       n = 3


Output: [1,2,2,3,5,6]
```

**Thoughts:** 

1. To fill the nums1 in place, should fill the array backwards: start with the last on both arrays and fill the larger one between nums\[i] and nums\[j]
2. **Code**

```cpp
class Solution {
public:
    void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {
        int i = m - 1, j = n - 1, tar = m + n - 1;
        while (j >= 0) {
            nums1[tar--] = i >= 0 && nums1[i] > nums2[j] ? nums1[i--] : nums2[j--];
        }
    }
};
```

Alternate version:

```java
class Solution {
    public void merge(int[] nums1, int m, int[] nums2, int n) {
        int i = m - 1, j = n - 1, merge = m + n - 1;
        // System.out.println("i: " + i + " j: " + j);
        while(i >= 0 && j >=0){
            nums1[merge--] = nums1[i] > nums2[j]?nums1[i--]:nums2[j--];
        }
        if (i< 0){
            // fill j
            while(j>=0) nums1[merge--] = nums2[j--];
        }
        if (j < 0){
            //fill i
            while(i>=0) nums1[merge--] = nums1[i--];
        }
        return;
    }
}
```