给定两个数组arr1 []和arr2 [] 。数组arr1 []已排序。任务是一一删除数组arr2 []中的每个元素，然后打印中位数的变化。
注意：数组arr2 []仅具有数组arr1 []中存在的那些元素。
例子：
Input: arr1[] = {2, 4, 6, 8, 10}, arr2[] = {4, 6}
Output: 1 1
Explanation:
Initially median is 6.
After removing 4, array becomes arr1[] = {2, 6, 8, 10}, median = 7, therefore the difference is 7 – 6 = 1.
After removing 6, array becomes arr1[] = {2, 8, 10}, median = 8, therefore the difference is 8 – 7 = 1.
Input: arr1[] = {1, 100, 250, 251}, arr2[] = {250, 1}
Output: -75 75.5
Explanation:
Initially median is 175.
After removing 250, array becomes arr1[] = {1, 100, 251}, median = 100, therefore the difference is 100 – 175 = -75.
After removing 1, array becomes arr1[] = {100, 251}, median = 175.5, therefore the difference is 175.5 – 100 = 75.5.
方法：我们的想法是遍历数组ARR2 []中的每个元素和从阵列ARR1 []删除每个元件和存储在一个阵列中的每个去除元件的后面的阵列ARR1 []的中值(比如说温度[])。从arr2 []中删除元素后，打印数组元素的连续差值以获取中位数的变化。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find the median change
// after removing elements from arr2[]
void medianChange(vector& arr1,
vector& arr2)
{
int N = arr1.size();
// To store the median
vector median;
// Store the current median
// If N is odd
if (N & 1) {
median
.push_back(arr1[N / 2] * 1.0);
}
// If N is even
else {
median
.push_back((arr1[N / 2]
+ arr1[(N - 1) / 2])
/ 2.0);
}
for (auto& x : arr2) {
// Find the current element
// in arr1
auto it = find(arr1.begin(),
arr1.end(),
x);
// Erase the element
arr1.erase(it);
// Decrement N
N--;
// Find the new median
// and append
// If N is odd
if (N & 1) {
median
.push_back(arr1[N / 2] * 1.0);
}
// If N is even
else {
median
.push_back((arr1[N / 2]
+ arr1[(N - 1) / 2])
/ 2.0);
}
}
// Print the corresponding
// difference of median
for (int i = 0;
i < median.size() - 1;
i++) {
cout << median[i + 1] - median[i]
<< ' ';
}
}
// Driven Code
int main()
{
// Given arrays
vector arr1 = { 2, 4, 6, 8, 10 };
vector arr2 = { 4, 6 };
// Function Call
medianChange(arr1, arr2);
return 0;
} Java
// Java program for the
// above approach
import java.util.*;
class GFG{
// Function to find the median
// change after removing elements
// from arr2[]
public static void medianChange(List arr1,
List arr2)
{
int N = arr1.size();
// To store the median
List median = new ArrayList<>();
// Store the current median
// If N is odd
if ((N & 1) != 0)
median.add(arr1.get(N / 2) * 1);
// If N is even
else
median.add((arr1.get(N / 2) +
arr1.get((N - 1) / 2)) / 2);
for(int x = 0; x < arr2.size(); x++)
{
// Find the current element
// in arr1
int it = arr1.indexOf(arr2.get(x));
// Erase the element
arr1.remove(it);
// Decrement N
N--;
// Find the new median
// and append
// If N is odd
if ((N & 1) != 0)
{
median.add(arr1.get(N / 2) * 1);
}
// If N is even
else
{
median.add((arr1.get(N / 2) +
arr1.get((N - 1) / 2)) / 2);
}
}
// Print the corresponding
// difference of median
for(int i = 0; i < median.size() - 1; i++)
{
System.out.print(median.get(i + 1) -
median.get(i) + " ");
}
}
// Driver Code
public static void main(String[] args)
{
// Given arrays
List arr1 = new ArrayList(){
{ add(2); add(4); add(6); add(8); add(10); } };
List arr2 = new ArrayList(){
{ add(4); add(6); } };
// Function Call
medianChange(arr1, arr2);
}
}
// This code is contributed by divyesh072019 Python3
# Python3 program for the
# above approach
# Function to find the median
# change after removing elements
# from arr2[]
def medianChange(arr1, arr2):
N = len(arr1)
# To store the median
median = []
# Store the current median
# If N is odd
if (N & 1):
median.append(arr1[N // 2] * 1)
# If N is even
else:
median.append((arr1[N // 2] +
arr1[(N - 1) // 2]) // 2)
for x in arr2:
# Find the current
# element in arr1
it = arr1.index(x)
# Erase the element
arr1.pop(it)
# Decrement N
N -= 1
# Find the new median
# and append
# If N is odd
if (N & 1):
median.append(arr1[N // 2] * 1)
# If N is even
else:
median.append((arr1[N // 2] +
arr1[(N - 1) // 2]) // 2)
# Print the corresponding
# difference of median
for i in range(len(median) - 1):
print(median[i + 1] - median[i],
end = ' ')
# Driver Code
if __name__ == "__main__":
# Given arrays
arr1 = [2, 4, 6,
8, 10]
arr2 = [4, 6]
# Function Call
medianChange(arr1, arr2)
# This code is contributed by ChitranayalC#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to find the median change
// after removing elements from arr2[]
static void medianChange(List arr1,
List arr2)
{
int N = arr1.Count;
// To store the median
List median = new List();
// Store the current median
// If N is odd
if ((N & 1) != 0)
{
median.Add(arr1[N / 2] * 1.0);
}
// If N is even
else
{
median.Add((arr1[N / 2] +
arr1[(N - 1) / 2]) / 2.0);
}
foreach(int x in arr2)
{
// Find the current element
// in arr1
int it = arr1.IndexOf(x);
// Erase the element
arr1.RemoveAt(it);
// Decrement N
N--;
// Find the new median
// and append
// If N is odd
if ((N & 1) != 0)
{
median.Add(arr1[N / 2] * 1.0);
}
// If N is even
else
{
median.Add((arr1[N / 2] +
arr1[(N - 1) / 2]) / 2.0);
}
}
// Print the corresponding
// difference of median
for(int i = 0; i < median.Count - 1; i++)
{
Console.Write(median[i + 1] -
median[i] + " ");
}
}
// Driver Code
static void Main()
{
// Given arrays
List arr1 = new List(
new int[]{ 2, 4, 6, 8, 10 });
List arr2 = new List(
new int[]{ 4, 6 });
// Function Call
medianChange(arr1, arr2);
}
}
// This code is contributed by divyeshrabadiya07 输出：
1 1时间复杂度： O(M * N)
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