> For the complete documentation index, see [llms.txt](https://dfine.gitbook.io/leetcode/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://dfine.gitbook.io/leetcode/88.merge_sorted_array.md).

# 88.Merge Sorted Array

**88.Merge Sorted Array**

难度: Easy

> 给定两个有序整数数组 nums1 和 nums2，将 nums2 合并到 nums1 中，使得 num1 成为一个有序数组。 说明: 初始化 nums1 和 nums2 的元素数量分别为 m 和 n。 你可以假设 nums1 有足够的空间（空间大小大于或等于 m + n）来保存 nums2 中的元素。

示例:

```
输入:
nums1 = [1,2,3,0,0,0], m = 3
nums2 = [2,5,6],       n = 3
输出: [1,2,2,3,5,6]
```

方法： 题目比较简单。最直观的就是新建一个nums1的长度的向量放最后合并的数组，不过会增加一些空间复杂度。 代码如下：

```
class Solution {
public:
    void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {
        vector<int> tmp=nums1;
        int N=nums1.size();
        int n1=0,n2=nums2.size();
        for(int i=N-1;i>=0;i--)
            if(nums1[i]){
                n1=i+1;
                break;
            }
        int s1=0,s2=0;
        while(s1<n1 && s2 <n2)
        {
            if(tmp[s1]<=nums2[s2])
            {
                nums1[s1+s2]=tmp[s1];
                ++s1;
            }
            else
            {
                nums1[s1+s2]=nums2[s2];
                ++s2;
            }
        }
        if(s1 == n1 && s2<n2)
            for(;s2<n2;++s2)
                nums1[s1+s2]=nums2[s2];
        if(s2==n2 && s1<n1)
            for(;s1<n1;++s1)
                nums1[s1+s2]=tmp[s1];
 
        
    }
};
```

这题并没有用到m和n，因为m和n本可以计算出来，即代码中的n1和n2。

另一种方法：直接原地操作，因为nums1后面空间较大，所以逆向合并即可。 代码如下：

```
class Solution {
public:
    void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {
        int s1=m-1,s2=n-1;
        for(int i=m+n-1;i>=0;--i)
        {
            if(s1>=0 && s2>=0)
            {
            if(nums1[s1]<nums2[s2])
            {
                nums1[i]=nums2[s2];
                --s2;
            }
            
            else
            {
                nums1[i]=nums1[s1];
                --s1;
            }
            }
            else 
                break;
        }
        if(s1<0 && s2<0)
            return ;
        else if(s1<0)
        {
            for (int i=s2;i>=0;i--)
            {
                nums1[i]=nums2[i];
            }
        }
 
        
    }
};
```


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