给定两个相同大小的数组 A[] 和 B[]（两个数组分别包含不同的元素，但可能有一些共同的元素），任务是形成相同大小的第三个（或结果）数组。生成的数组应具有来自两个数组的最多 n 个元素。它应该首先选择 A[] 的元素，然后按照它们在原始数组中出现的相同顺序选择 B[] 的元素。如果有共同元素，那么 res[] 中应该只有一个元素，并且应该优先考虑 A[]。
例子：
Input : A[] = [ 9 7 2 3 6 ]
B[] = [ 7 4 8 0 1 ]
Output : res[] = [9 7 6 4 8]
res[] has maximum n elements of both A[]
and B[] such that elements of A[] appear
first (in same order), then elements of B[].
Also 7 is common and priority is given to
A's 7.
Input : A[] = [ 6 7 5 3 ]
B[] = [ 5 6 2 9 ]
Output : res[] = [ 6 7 5 9 ] 1) 创建两个数组的副本并按降序对副本进行排序。
2) 使用哈希选择两个数组中唯一的 n 个最大元素，优先考虑 A[]。
3) 将结果数组初始化为空。
4) 遍历 A[]，复制 A[] 中存在于哈希中的那些元素。这样做是为了保持元素的顺序相同。
5) 对 B[] 重复步骤 4。这次我们只考虑那些不在 A[] 中的元素（不要在哈希中出现两次）。
下面是上述想法的实现。
C++
// Make a set of maximum elements from two
// arrays A[] and B[]
#include
using namespace std;
void maximizeTheFirstArray(int A[], int B[],
int n)
{
// Create copies of A[] and B[] and sort
// the copies in descending order.
vector temp1(A, A+n);
vector temp2(B, B+n);
sort(temp1.begin(), temp1.end(), greater());
sort(temp2.begin(), temp2.end(), greater());
// Put maximum n distinct elements of
// both sorted arrays in a map.
unordered_map m;
int i = 0, j = 0;
while (m.size() < n)
{
if (temp1[i] >= temp2[j])
{
m[temp1[i]]++;
i++;
}
else
{
m[temp2[j]]++;
j++;
}
}
// Copy elements of A[] to that
// are present in hash m.
vector res;
for (int i = 0; i < n; i++)
if (m.find(A[i]) != m.end())
res.push_back(A[i]);
// Copy elements of B[] to that
// are present in hash m. This time
// we also check if the element did
// not appear twice.
for (int i = 0; i < n; i++)
if (m.find(B[i]) != m.end() &&
m[B[i]] == 1)
res.push_back(B[i]);
// print result
for (int i = 0; i < n; i++)
cout << res[i] << " ";
}
// driver program
int main()
{
int A[] = { 9, 7, 2, 3, 6 };
int B[] = { 7, 4, 8, 0, 1 };
int n = sizeof(A) / sizeof(A[0]);
maximizeTheFirstArray(A, B, n);
return 0;
} Java
// Make a set of maximum elements from two
// arrays A[] and B[]
import java.io.*;
import java.util.*;
class GFG
{
static void maximizeTheFirstArray(int[] A, int[] B,int n)
{
// Create copies of A[] and B[] and sort
// the copies in descending order.
ArrayList temp1 = new ArrayList();
ArrayList temp2 = new ArrayList();
for(int i : A)
{
temp1.add(i);
}
for(int i:B)
{
temp2.add(i);
}
Collections.sort(temp1, Collections.reverseOrder());
Collections.sort(temp2, Collections.reverseOrder());
// Put maximum n distinct elements of
// both sorted arrays in a map.
Map m = new HashMap<>();
int i = 0, j = 0;
while (m.size() < n)
{
if (temp1.get(i) >= temp2.get(j))
{
if(m.containsKey(temp1.get(i)))
{
m.put(temp1.get(i), m.get(temp1.get(i)) + 1);
}
else
{
m.put(temp1.get(i), 1);
}
i++;
}
else
{
if(m.containsKey(temp2.get(j)))
{
m.put(temp2.get(j), m.get(temp2.get(j)) + 1);
}
else
{
m.put(temp2.get(j), 1);
}
j++;
}
}
// Copy elements of A[] to that
// are present in hash m.
ArrayList res = new ArrayList();
for (i = 0; i < n; i++)
if (m.containsKey(A[i]))
res.add(A[i]);
// Copy elements of B[] to that
// are present in hash m. This time
// we also check if the element did
// not appear twice.
for (i = 0; i < n; i++)
if (m.containsKey(B[i]) && m.get(B[i]) == 1)
res.add(B[i]);
// print result
for (i = 0; i < n; i++)
System.out.print(res.get(i)+" ");
}
// Driver program
public static void main (String[] args)
{
int A[] = { 9, 7, 2, 3, 6 };
int B[] = { 7, 4, 8, 0, 1 };
int n = A.length;
maximizeTheFirstArray(A, B, n);
}
}
// This code is contributed by rag2127 Python3
# Python3 program to implement the
# above approach
# Make a set of maximum elements
# from two arrays A[] and B[]
from collections import defaultdict
def maximizeTheFirstArray(A, B, n):
# Create copies of A[] and B[]
# and sort the copies in
# descending order.
temp1 = A.copy()
temp2 = B.copy()
temp1.sort(reverse = True)
temp2.sort(reverse = True)
# Put maximum n distinct
# elements of both sorted
# arrays in a map.
m = defaultdict(int)
i = 0
j = 0;
while (len(m) < n):
if (temp1[i] >= temp2[j]):
m[temp1[i]] += 1
i += 1
else:
m[temp2[j]] += 1
j += 1
# Copy elements of A[] to that
# are present in hash m.
res = []
for i in range (n):
if (A[i] in m):
res.append(A[i])
# Copy elements of B[] to that
# are present in hash m. This time
# we also check if the element did
# not appear twice.
for i in range (n):
if (B[i] in m and
m[B[i]] == 1):
res.append(B[i])
# Print result
for i in range (n):
print (res[i], end = " ")
# Driver code
if __name__ == "__main__":
A = [9, 7, 2, 3, 6]
B = [7, 4, 8, 0, 1]
n = len(A)
maximizeTheFirstArray(A, B, n);
# This code is contributed by ChitranayalC#
// Make a set of maximum elements from two
// arrays A[] and B[]
using System;
using System.Collections.Generic;
class GFG{
static void maximizeTheFirstArray(int[] A, int[] B,
int n)
{
// Create copies of A[] and B[] and sort
// the copies in descending order.
List temp1 = new List();
List temp2 = new List();
foreach(int i in A)
{
temp1.Add(i);
}
foreach(int i in B)
{
temp2.Add(i);
}
temp1.Sort();
temp1.Reverse();
temp2.Sort();
temp2.Reverse();
// Put maximum n distinct elements of
// both sorted arrays in a map.
Dictionary m = new Dictionary();
int I = 0, j = 0;
while (m.Count < n)
{
if (temp1[I] >= temp2[j])
{
if (m.ContainsKey(temp1[I]))
{
m[temp1[I]]++;
}
else
{
m.Add(temp1[I], 1);
}
I++;
}
else
{
if (m.ContainsKey(temp2[j]))
{
m[temp2[j]]++;
}
else
{
m.Add(temp2[j], 1);
}
j++;
}
}
// Copy elements of A[] to that
// are present in hash m.
List res = new List();
for(int i = 0; i < n; i++)
if (m.ContainsKey(A[i]))
res.Add(A[i]);
// Copy elements of B[] to that
// are present in hash m. This time
// we also check if the element did
// not appear twice.
for(int i = 0; i < n; i++)
if (m.ContainsKey(B[i]) && m[B[i]] == 1)
res.Add(B[i]);
// print result
for(int i = 0; i < n; i++)
Console.Write(res[i] + " ");
}
// Driver Code
static public void Main()
{
int[] A = { 9, 7, 2, 3, 6 };
int[] B = { 7, 4, 8, 0, 1 };
int n = A.Length;
maximizeTheFirstArray(A, B, n);
}
}
// This code is contributed by avanitrachhadiya2155 Javascript
输出：
9 7 6 4 8 时间复杂度： O(n Log n)
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