给定两个已排序的数组，大小为N和M 的arr[] 、 brr[] ，任务是合并两个给定数组，使它们形成组合两个数组元素的整数排序序列。
例子：
Input: arr[] = {10}, brr[] = {2, 3}
Output: 2 3 10
Explanation: The merged sorted array obtained by taking all the elements from the both the arrays is {2, 3, 10}.
Therefore, the required output is 2 3 10.
Input: arr[] = {1, 5, 9, 10, 15, 20}, brr[] = {2, 3, 8, 13}
Output: 1 2 3 5 8 9 10 13 15 20
朴素的方法：请参阅合并两个已排序的数组，了解合并两个给定数组的最简单方法。
时间复杂度： O(N * M)
辅助空间： O(1)
空间优化方法：请参阅合并两个已排序数组与 O(1) 额外空间以合并两个给定数组而不使用任何额外内存。
时间复杂度： O(N * M)
辅助空间： O(1)
Efficient Space Optimized Approach：参考Efficiently merging两个已排序数组与O(1)额外空间来合并两个给定数组而不使用任何额外内存。
时间复杂度： O((N + M) * log(N + M))
辅助空间： O(1)
基于分区的方法：想法是将最终排序数组的第(N + 1)个元素视为主元元素，并围绕主元元素执行快速排序分区。最后，将最终排序数组的前N 个较小元素存储到数组arr[] 中，并将最终排序数组的最后M 个较大元素以任意顺序存储到数组brr[]中，并分别对这两个数组进行排序。请按照以下步骤解决问题：
- 初始化一个变量，比如index来存储最终排序数组的每个元素的索引。
- 找到第(N + 1)个元素 最终排序的数组作为枢轴元素。
- 围绕枢轴元素执行快速排序分区。
- 最后，分别对数组arr[]和brr[] 进行排序。
下面是上述方法的实现：
C++
// C++ program to implement
// the above approach
#include
using namespace std;
// Function to perform the partition
// around the pivot element
void partition(int arr[], int N,
int brr[], int M,
int Pivot)
{
// Stores index of each element
// of the array, arr[]
int l = N - 1;
// Stores index of each element
// of the array, brr[]
int r = 0;
// Traverse both the array
while (l >= 0 && r < M) {
// If pivot is
// smaller than arr[l]
if (arr[l] < Pivot)
l--;
// If Pivot is
// greater than brr[r]
else if (brr[r] > Pivot)
r++;
// If either arr[l] > Pivot
// or brr[r] < Pivot
else {
swap(arr[l], brr[r]);
l--;
r++;
}
}
}
// Function to merge
// the two sorted array
void Merge(int arr[], int N,
int brr[], int M)
{
// Stores index of each element
// of the array arr[]
int l = 0;
// Stores index of each element
// of the array brr[]
int r = 0;
// Stores index of each element
// the final sorted array
int index = -1;
// Stores the pivot element
int Pivot = 0;
// Traverse both the array
while (index < N && l < N && r < M) {
if (arr[l] < brr[r]) {
Pivot = arr[l++];
}
else {
Pivot = brr[r++];
}
index++;
}
// If pivot element is not found
// or index < N
while (index < N && l < N) {
Pivot = arr[l++];
index++;
}
// If pivot element is not found
// or index < N
while (index < N && r < M) {
Pivot = brr[r++];
index++;
}
// Place the first N elements of
// the sorted array into arr[]
// and the last M elements of
// the sorted array into brr[]
partition(arr, N, brr,
M, Pivot);
// Sort both the arrays
sort(arr, arr + N);
sort(brr, brr + M);
// Print the first N elements
// in sorted order
for (int i = 0; i < N; i++)
cout << arr[i] << " ";
// Print the last M elements
// in sorted order
for (int i = 0; i < M; i++)
cout << brr[i] << " ";
}
// Driver Code
int main()
{
int arr[] = { 1, 5, 9 };
int brr[] = { 2, 4, 7, 10 };
int N = sizeof(arr) / sizeof(arr[0]);
int M = sizeof(brr) / sizeof(brr[0]);
Merge(arr, N, brr, M);
return 0;
} Java
// Java program to implement
// the above approach
import java.util.*;
class GFG{
// Function to perform the partition
// around the pivot element
static void partition(int arr[], int N,
int brr[], int M,
int Pivot)
{
// Stores index of each element
// of the array, arr[]
int l = N - 1;
// Stores index of each element
// of the array, brr[]
int r = 0;
// Traverse both the array
while (l >= 0 && r < M)
{
// If pivot is
// smaller than arr[l]
if (arr[l] < Pivot)
l--;
// If Pivot is
// greater than brr[r]
else if (brr[r] > Pivot)
r++;
// If either arr[l] > Pivot
// or brr[r] < Pivot
else
{
int t = arr[l];
arr[l] = brr[r];
brr[r] = t;
l--;
r++;
}
}
}
// Function to merge
// the two sorted array
static void Merge(int arr[], int N,
int brr[], int M)
{
// Stores index of each element
// of the array arr[]
int l = 0;
// Stores index of each element
// of the array brr[]
int r = 0;
// Stores index of each element
// the final sorted array
int index = -1;
// Stores the pivot element
int Pivot = 0;
// Traverse both the array
while (index < N && l < N &&
r < M)
{
if (arr[l] < brr[r])
{
Pivot = arr[l++];
}
else
{
Pivot = brr[r++];
}
index++;
}
// If pivot element is not found
// or index < N
while (index < N && l < N)
{
Pivot = arr[l++];
index++;
}
// If pivot element is not
// found or index < N
while (index < N && r < M)
{
Pivot = brr[r++];
index++;
}
// Place the first N elements of
// the sorted array into arr[]
// and the last M elements of
// the sorted array into brr[]
partition(arr, N, brr,
M, Pivot);
// Sort both the arrays
Arrays.sort(arr);
Arrays.sort(brr);
// Print the first N elements
// in sorted order
for (int i = 0; i < N; i++)
System.out.print(arr[i] + " ");
// Print the last M elements
// in sorted order
for (int i = 0; i < M; i++)
System.out.print(brr[i] + " ");
}
// Driver Code
public static void main(String[] args)
{
int arr[] = {1, 5, 9};
int brr[] = {2, 4, 7, 10};
int N = arr.length;
int M = brr.length;
Merge(arr, N, brr, M);
}
}
// This code is contributed by Amit KatiyarPython3
# Python3 program to implement
# the above approach
# Function to perform the partition
# around the pivot element
def partition(arr, N, brr, M, Pivot):
# Stores index of each element
# of the array, arr[]
l = N - 1
# Stores index of each element
# of the array, brr[]
r = 0
# Traverse both the array
while (l >= 0 and r < M):
# If pivot is smaller
# than arr[l]
if (arr[l] < Pivot):
l -= 1
# If Pivot is greater
# than brr[r]
elif (brr[r] > Pivot):
r += 1
# If either arr[l] > Pivot
# or brr[r] < Pivot
else:
arr[l], brr[r] = brr[r], arr[l]
l -= 1
r += 1
# Function to merge
# the two sorted array
def Merge(arr, N, brr, M):
# Stores index of each element
# of the array arr[]
l = 0
# Stores index of each element
# of the array brr[]
r = 0
# Stores index of each element
# the final sorted array
index = -1
# Stores the pivot element
Pivot = 0
# Traverse both the array
while (index < N and l < N and r < M):
if (arr[l] < brr[r]):
Pivot = arr[l]
l += 1
else:
Pivot = brr[r]
r += 1
index += 1
# If pivot element is not found
# or index < N
while (index < N and l < N):
Pivot = arr[l]
l += 1
index += 1
# If pivot element is not found
# or index < N
while (index < N and r < M):
Pivot = brr[r]
r += 1
index += 1
# Place the first N elements of
# the sorted array into arr[]
# and the last M elements of
# the sorted array into brr[]
partition(arr, N, brr, M, Pivot)
# Sort both the arrays
arr = sorted(arr)
brr = sorted(brr)
# Print the first N elements
# in sorted order
for i in range(N):
print(arr[i], end = " ")
# Print the last M elements
# in sorted order
for i in range(M):
print(brr[i], end = " ")
# Driver Code
if __name__ == '__main__':
arr = [ 1, 5, 9 ]
brr = [ 2, 4, 7, 10 ]
N = len(arr)
M = len(brr)
Merge(arr, N, brr, M)
# This code is contributed by mohit kumar 29C#
// C# program to implement
// the above approach
using System;
class GFG{
// Function to perform the
// partition around the pivot
// element
static void partition(int []arr, int N,
int []brr, int M,
int Pivot)
{
// Stores index of each element
// of the array, []arr
int l = N - 1;
// Stores index of each element
// of the array, brr[]
int r = 0;
// Traverse both the array
while (l >= 0 && r < M)
{
// If pivot is
// smaller than arr[l]
if (arr[l] < Pivot)
l--;
// If Pivot is
// greater than brr[r]
else if (brr[r] > Pivot)
r++;
// If either arr[l] > Pivot
// or brr[r] < Pivot
else
{
int t = arr[l];
arr[l] = brr[r];
brr[r] = t;
l--;
r++;
}
}
}
// Function to merge
// the two sorted array
static void Merge(int []arr, int N,
int []brr, int M)
{
// Stores index of each element
// of the array []arr
int l = 0;
// Stores index of each element
// of the array brr[]
int r = 0;
// Stores index of each element
// the readonly sorted array
int index = -1;
// Stores the pivot element
int Pivot = 0;
// Traverse both the array
while (index < N && l < N &&
r < M)
{
if (arr[l] < brr[r])
{
Pivot = arr[l++];
}
else
{
Pivot = brr[r++];
}
index++;
}
// If pivot element is not found
// or index < N
while (index < N && l < N)
{
Pivot = arr[l++];
index++;
}
// If pivot element is not
// found or index < N
while (index < N && r < M)
{
Pivot = brr[r++];
index++;
}
// Place the first N elements of
// the sorted array into []arr
// and the last M elements of
// the sorted array into brr[]
partition(arr, N, brr,
M, Pivot);
// Sort both the arrays
Array.Sort(arr);
Array.Sort(brr);
// Print the first N elements
// in sorted order
for (int i = 0; i < N; i++)
Console.Write(arr[i] + " ");
// Print the last M elements
// in sorted order
for (int i = 0; i < M; i++)
Console.Write(brr[i] + " ");
}
// Driver Code
public static void Main(String[] args)
{
int []arr = {1, 5, 9};
int []brr= {2, 4, 7, 10};
int N = arr.Length;
int M = brr.Length;
Merge(arr, N, brr, M);
}
}
// This code is contributed by shikhasingrajput输出
1 2 4 5 7 9 10 时间复杂度： O((N + M)log(N + M))
辅助空间： O(1)
高效的方法：参考合并两个已排序的数组来有效地合并两个给定的数组。
时间复杂度： O(N + M)
辅助空间： O(N + M)
如果您想与行业专家一起参加直播课程，请参阅Geeks Classes Live