> For the complete documentation index, see [llms.txt](https://celia-qian.gitbook.io/leetcode-notebook-2020-2021/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://celia-qian.gitbook.io/leetcode-notebook-2020-2021/linear-list/array/median-of-two-sorted-arrays.md).

# Median of Two Sorted Arrays

## Median of Two Sorted Arrays

原题链接：<https://leetcode.com/problems/median-of-two-sorted-arrays/>

### 描述

> Given two sorted arrays `nums1` and `nums2` of size `m` and `n` respectively, return **the median** of the two sorted arrays.

### 例子

> Input: nums1 = \[1,3], nums2 = \[2]&#x20;
>
> Output: 2.00000&#x20;
>
> Explanation: merged array = \[1,2,3] and median is 2.
>
> Input: nums1 = \[1,2], nums2 = \[3,4]&#x20;
>
> Output: 2.50000&#x20;
>
> Explanation: merged array = \[1,2,3,4] and median is (2 + 3) / 2 = 2.5.
>
> Input: nums1 = \[0,0], nums2 = \[0,0]&#x20;
>
> Output: 0.00000
>
> Input: nums1 = \[], nums2 = \[1]&#x20;
>
> Output: 1.00000

**Follow up:**&#x20;

The overall run time complexity should be `O(log (m+n))`.<br>

**Constrain：**

> * `nums1.length == m`
> * `nums2.length == n`
> * `0 <= m <= 1000`
> * `0 <= n <= 1000`
> * `1 <= m + n <= 2000`
> * `-106 <= nums1[i], nums2[i] <= 106`

### 解法一

&#x20;      O(m+n)的解法比较直观，直接merge两个数组，然后找到第k大的元素（题中及寻找中位数） 但是merge之后，如果使用“排序”的操作，时间花费是expensive的。&#x20;

&#x20;      也可以用一个计数器，记录当前已经找到第m大的元素，同时再使用两个指针pA和pB，分别指向A和B数组的第一个元素，使用merge sort的原理：&#x20;

&#x20;      如果数组A当前元素小，则pA++，同时m++；&#x20;

&#x20;      如果数组B当前元素小，则pB++，同时m++。

&#x20;      最终当m等于k的时候，及得到答案，时间复杂度：O（K），空间复杂度O（1） 如果当K很接近m+n的时候，这个方法还是O（m+n）的。

```cpp
class Solution {
public:
    double findMedianSortedArrays(vector<int>& A, vector<int>& B) {
        int m = A.size();
        int n = B.size();
        if(m > n){
            vector<int> temp; 
            temp.assign(A.begin(),A.end());
            A.assign(B.begin(),B.end());
            B.assign(temp.begin(),temp.end());
            int tmp = m; m = n; n = tmp;
        }
        int imin = 0, imax = m, halflen = (m+n+1)/2;
        while(imin <= imax){
            int i = (imin+imax)/2;
            int j = halflen-i;
            if(i < imax && B[j-1]>A[i]){
                imin = i + 1; // i is too small
            }
            else if(i > imin && A[i-1] > B[j]){
                imax = i - 1; // i is too big
            }else{
                int maxleft = 0;
                if(i == 0){maxleft = B[j-1];}
                else if(j == 0){maxleft = A[i-1];}
                else {maxleft = max(A[i-1], B[j-1]);}
                if((m + n) % 2 == 1){return maxleft;}
                
                int minright = 0;
                if(i == m){minright = B[j];}
                else if(j == n){minright = A[i];}
                else {minright = min(A[i], B[j]);}
                
                return (maxleft + minright) / 2.0;
            }
        }
        return 0.0;
    }
};
```

### 解法二

递归的方法：

```cpp
class Solution {
public:
    double findMedianSortedArrays(const vector<int>& A, const vector<int>& B) {
        const int m = A.size();
        const int n = B.size();
        int total = m + n;
        if (total & 0x1)
            return find_kth(A.begin(), m, B.begin(), n, total / 2 + 1);
        else
            return (find_kth(A.begin(), m, B.begin(), n, total / 2) + find_kth(A.begin(), m, B.begin(), n, total / 2 + 1)) / 2.0;
    } 
    private:
    static int find_kth(std::vector<int>::const_iterator A, int m, std::vector<int>::const_iterator B, int n, int k) {
        //always assume that m is equal or smaller than n 
        if (m > n) return find_kth(B, n, A, m, k);
        if (m == 0) return *(B + k - 1);
        if (k == 1) return min(*A, *B);
        //divide k into two parts
        int ia = min(k / 2, m), ib = k - ia; 
        if (*(A + ia - 1) < *(B + ib - 1))
            return find_kth(A + ia, m - ia, B, n, k - ia); 
        else if (*(A + ia - 1) > *(B + ib - 1))
            return find_kth(A, m, B + ib, n - ib, k - ib); 
        else
            return A[ia - 1];
        } 
};

```

## 总结