两个排序数组的中位数
中位数是中间数，换句话说，中位数是有序列表中的中间观察值。它对应于50％的累积百分比。
两个数组的大小必须相同，我们首先会找到两个单独数组的中值，然后比较这些单独的中值以获得两个列表的实际中值。
输入输出
Input:
Two sorted array are given.
Array 1: {1, 2, 3, 6, 7}
Array 2: {4, 6, 8, 10, 11}
Output:
The median from two array. Here the median value is 6.
Merge the given lists into one. {1, 2, 3, 4, 6, 6, 7, 8, 10, 11}
From the merged list find the average of two middle elements. here (6+6)/2 = 6.算法
中位数（列表，n）
输入： 数据列表和数据数。
输出： 给定列表的中位数。
Begin
if the list has even number of data, then
return (list[n/2] + list[n/2-1])/2
else
return list[n/2]
EndfindMedian(list1,list2,n)
输入-两个排序列表，以及列表数。
输出- 来自两个排序列表的中位数。
Begin
if n <= 0, then
it is invalid, and return invalid number
if n = 1, then
return (list1[0] + list2[0])/2
if n = 2, then
return ((max of list1[0], list2[0]) + (min of list1[1], list2[1]))/2
med1 := median(list1, n)
med2 := median(list2, n)
if med1 = med2, then
return med1
if med1 < med2, then
if item has even number of data, then
subList := data from list2, from 0 to n/2 – 1 data
return findMedian(subList, list1, n – (n/2) + 1)
subList := data from list2, from 0 to n/2 data
return findMedian(subList, list2, n – (n/2))
End示例
#include<iostream>
using namespace std;
int median(int list[], int n) {
if (n%2 == 0) //when array containts even number of data
return (list[n/2] + list[n/2-1])/2;
else //for odd number of data
return list[n/2];
}
intfindMedian(int list1[], int list2[], int n) {
if (n <= 0)
return -1; //invalid length of lists
if (n == 1)
return (list1[0] + list2[0])/2; //for single element simply get average from two array
if (n == 2)
return (max(list1[0], list2[0]) + min(list1[1], list2[1])) / 2;
int med1 = median(list1, n); //Find median from first array
int med2 = median(list2, n); //Find median from second array
if (med1 == med2) //when both medians are same, they are the final median
return med1;
if (med1 < med2) {
if (n % 2 == 0)
return findMedian(list1 + n/2 - 1, list2, n - n/2 +1);
return findMedian(list1 + n/2, list2, n - n/2);
}
if (n % 2 == 0) //when med1 > med2
return findMedian(list2 + n/2 - 1, list1, n - n/2 + 1);
return findMedian(list2 + n/2, list1, n - n/2);
}
int main() {
int list1[] = {1, 2, 3, 6, 7};
int list2[] = {4, 6, 8, 10, 11};
int n1 = 5;
int n2 = 5;
if (n1 == n2)
cout<< "Median is "<<findMedian(list1, list2, n1);
else
cout<< "Doesn't work for lists of unequal size";
}输出结果
Median is 6