LeetCode刷題：4. Median of Two Sorted Arrays

題目：

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！

代碼：

/*
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;

// 定義排序比較函數
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一個新的整數向量a，將nums1和nums2都放進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];

		// 對a使用標準庫算法qsort（）排序
		qsort(a, nums1.size() + nums2.size(), sizeof(int), comp_my);

		//計算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;
	}
};