LeetCode刷题：4. Median of Two Sorted Arrays

题目：There are two sorted arrays nums1 and nums2 of size m and n respectively.Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).Example 1:
nums1 = [1, 3]
nums2 = [2]

The median is 2.0
Example 2:
nums1 = [1, 2]
nums2 = [3, 4]

The median is (2 + 3)/2 = 2.5

思路：
暴力排序！

代码：/*
There are two sorted arrays nums1 and nums2 of size m and n respectively.
Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).

Example 1:
nums1 = [1, 3]
nums2 = [2]
The median is 2.0

Example 2:
nums1 = [1, 2]
nums2 = [3, 4]
The median is (2 + 3)/2 = 2.5
*/

#include <vector>

using namespace std;

// 定义排序比较函数
int comp_my(const void* a, const void* b)
{
return *(int*)a - *(int*)b;
}

class Solution {
public:
double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
// new一个新的整数向量a，将nums1和nums2都放进a中
int *a = new int[nums1.size() + nums2.size()];
for (int i = 0; i < nums1.size(); i++)
a[i] = nums1[i];
for (int i = 0; i < nums2.size(); i++)
a[nums1.size() + i] = nums2[i];

// 对a使用标准库算法qsort（）排序
qsort(a, nums1.size() + nums2.size(), sizeof(int), comp_my);

//计算medium值
int nums_tol = nums1.size() + nums2.size();
double median = (nums_tol % 2) ? a[nums_tol / 2] : (a[nums_tol / 2] + a[nums_tol / 2 - 1]) / 2.0;

return median;
}
};