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法学两个排序数组的中值
中值是中间数，换句话说，中值是已排序列表中的中间观测值。它对应于 50% 的累积百分比。
两个数组的大小必须相等，我们首先将找到两个单独数组的中值，然后比较单独的中值以获得两个列表的实际中值。
输入和输出
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
median(list, n)
输入：数据列表及数据数量。
输出：给定列表的中值。
Begin if the list has even number of data, then return (list[n/2] + list[n/2-1])/2 else return list[n/2] End
findMedian(list1, list2, n)
输入 − 两个排序列表以及列表数。
输出 − 两个排序列表中的中值。
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
Median is 6
更新于： 16-Jun-2020
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