# Merge Sorted Array

**Easy**

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]
```

## Analysis

这一题的trick在于，因为是数组，不适合插入操作，所以从nums1\[]的末端开始反向填充，就可以避免数据大量搬运。

## Solution

One pass - O(n) time, O(1) space - (2ms, 100%)

```java
class Solution {
    public void merge(int[] nums1, int m, int[] nums2, int n) {
        int k = m + n;
        int i = m - 1;
        int j = n - 1;
        for (int l = k - 1; l >= 0; l--) {
            // nums2[] elements have all been copied
            if (j < 0) {
                return;
            }
            // nums1[] elements have all been copied
            if (i < 0) {
                nums1[l] = nums2[j];
                j--;
                continue;
            }

            // compare and assign elements from nums1, nums2
            if (nums2[j] > nums1[i]) {
                nums1[l] = nums2[j];
                j--;
            } else {
                nums1[l] = nums1[i];
                i--;
            }
        }

    }
}


// [1,2,3,0,0,0]
// [4,5,6]

// [4,5,6,0,0,0]
// [1,2,3]
```

Another more concise implementation (2ms, 100%)

```java
class Solution {

    public void merge(int[] nums1, int m, int[] nums2, int n) {
        int k = m + n - 1;
        int i = m - 1;
        int j = n - 1;
        while (i >= 0 && j>= 0) {
            // compare and assign elements from nums1, nums2
            if (nums2[j] > nums1[i]) {
                nums1[k--] = nums2[j--];
            } else {
                nums1[k--] = nums1[i--];
            }
        }
        while (j >= 0) {
            nums1[k--] = nums2[j--];
        }
    }
}
```


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