Description:
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)).
You may assume nums1 and nums2 cannot be both empty.
Example1:
nums1 = [1, 3]
nums2 = [2]
Example2:
nums1 = [1, 2]
nums2 = [3, 4]
The median is (2 + 3)/2 = 2.5
解题方法:
merge sort(归并排序)
double findMedianSortedArrays(int* nums1, int nums1Size, int* nums2, int nums2Size){
int* nums = (int*)malloc((nums1Size + nums2Size) * sizeof(int));
double m;
int i = 0, j = 0, k = 0;
while (i < nums1Size&&j < nums2Size)
{
if (nums1[i] < nums2[j])
{
nums[k++] = nums1[i++];
}
else
{
nums[k++] = nums2[j++];
}
}
//nums1还有一些数
while (i < nums1Size)
nums[k++] = nums1[i++];
//nums2还有一些数
while (j < nums2Size)
nums[k++] = nums2[j++];
//找到中位数并返回
if ((nums1Size + nums2Size) % 2 == 0)
{
int pos = (nums1Size + nums2Size) / 2;
m = (nums[pos] + nums[pos - 1]) / 2.0;
return m;
}
else
{
int pos = (nums1Size + nums2Size) / 2;
m = nums[pos];
return m;
}
}
运行结果: