合并K个排序的数组|第三组(使用分而治之方法)

给定k个排序数组，每个数组的大小为N ，任务是将它们合并为单个排序数组。
例子：

Input: arr[][] = {{5, 7, 15, 18},
                   {1, 8, 9, 17},
                   {1, 4, 7, 7}}
Output: {1, 1, 4, 5, 7, 7, 7, 8, 9, 15, 17, 18}

Input: arr[][] = {{3, 2, 1}
                   {6, 5, 4}}
Output:  {1, 2, 3, 4, 5, 6}

先决条件：合并排序
简单方法：一个简单的解决方案是将所有数组一个接一个地追加并排序。

CPP14

// C++ program to merge K
// sorted arrays
#include 
using namespace std;
#define N 4
 
// Merge and sort k arrays
void merge_and_sort(
    int* output, int arr[][N],
    int n, int k)
{
    // Put the elments in sorted array.
    for (int i = 0; i < k; i++) {
        for (int j = 0; j < n; j++) {
            output[i * n + j] = arr[i][j];
        }
    }
 
    // Sort the output array
    sort(output, output + n * k);
}
 
// Driver Function
int main()
{
    // Input 2D-array
    int arr[][N] = { { 5, 7, 15, 18 },
                     { 1, 8, 9, 17 },
                     { 1, 4, 7, 7 } };
    int n = N;
 
    // Number of arrays
    int k = sizeof(arr) / sizeof(arr[0]);
 
    // Output array
    int* output = new int[n * k];
 
    merge_and_sort(output, arr, n, k);
 
    // Print merged array
    for (int i = 0; i < n * k; i++)
        cout << output[i] << " ";
 
    return 0;
}

Java

// Java program to merge K
// sorted arrays
import java.util.*;
class GFG
{
    static int N = 4; 
     
    // Merge and sort k arrays
    static void merge_and_sort(int output[], int arr[][],
                               int n, int k)
    {
        // Put the elments in sorted array.
        for (int i = 0; i < k; i++)
        {
            for (int j = 0; j < n; j++)
            {
                output[i * n + j] = arr[i][j];
            }
        }
       
        // Sort the output array
        Arrays.sort(output);
    }
 
  // Driver code
    public static void main(String[] args)
    {
       
        // Input 2D-array
        int arr[][] = { { 5, 7, 15, 18 },
                         { 1, 8, 9, 17 },
                         { 1, 4, 7, 7 } };
        int n = N;
       
        // Number of arrays
        int k = arr.length;
       
        // Output array
        int output[] = new int[n * k];     
        merge_and_sort(output, arr, n, k);
       
        // Print merged array
        for (int i = 0; i < n * k; i++)
            System.out.print(output[i] + " ");
    }
}
 
// This code is contributed by divyesh072019

Python3

# Python3 program to merge K
# sorted arrays
N = 4
 
# Merge and sort k arrays
def merge_and_sort(output, arr, n, k):
 
    # Put the elments in sorted array.
    for  i in range(k):
        for j in range(n):
            output[i * n + j] = arr[i][j];
 
    # Sort the output array
    output.sort()
 
# Driver Function
if __name__=='__main__':
 
    # Input 2D-array
    arr = [ [ 5, 7, 15, 18 ],
                     [ 1, 8, 9, 17 ],
                     [ 1, 4, 7, 7 ] ];
    n = N;
 
    # Number of arrays
    k = len(arr)
 
    # Output array
    output = [0 for i in range(n * k)]
    merge_and_sort(output, arr, n, k);
 
    # Print merged array
    for  i in range(n * k):
        print(output[i], end = ' ')
     
# This code is contributed by rutvik_56.

C#

// C# program to merge K
// sorted arrays
using System;
class GFG
{
     
    static int N = 4; 
     
    // Merge and sort k arrays
    static void merge_and_sort(int[] output, int[,] arr,
                               int n, int k)
    {
        // Put the elments in sorted array.
        for (int i = 0; i < k; i++)
        {
            for (int j = 0; j < n; j++)
            {
                output[i * n + j] = arr[i,j];
            }
        }
       
        // Sort the output array
        Array.Sort(output);
    }
   
  // Driver code
  static void Main()
  {
     
    // Input 2D-array
    int[,] arr = { { 5, 7, 15, 18 },
                     { 1, 8, 9, 17 },
                     { 1, 4, 7, 7 } };
    int n = N;
   
    // Number of arrays
    int k = arr.GetLength(0);
   
    // Output array
    int[] output = new int[n * k];     
    merge_and_sort(output, arr, n, k);
   
    // Print merged array
    for (int i = 0; i < n * k; i++)
        Console.Write(output[i] + " ");
  }
}
 
// This code is contributed by divyeshrabadiya07

Javascript

    // Javascript program to merge K
    // sorted arrays
     
    let N = 4;
      
    // Merge and sort k arrays
    function merge_and_sort(output, arr, n, k)
    {
        // Put the elments in sorted array.
        for (let i = 0; i < k; i++)
        {
            for (let j = 0; j < n; j++)
            {
                output[i * n + j] = arr[i][j];
            }
        }
        
        // Sort the output array
        output.sort(function(a, b){return a - b});
    }
     
    // Input 2D-array
    let arr = [ [ 5, 7, 15, 18 ],
                 [ 1, 8, 9, 17 ],
                 [ 1, 4, 7, 7 ] ];
    let n = N;
    
    // Number of arrays
    let k = 3;
    
    // Output array
    let output = new Array(n * k);    
    merge_and_sort(output, arr, n, k);
    
    // Print merged array
    for (let i = 0; i < n * k; i++)
        document.write(output[i] + " ");
         
        // This code is contributed by mukesh07.

输出：

1 1 4 5 7 7 7 8 9 15 17 18

复杂度分析：

• 时间复杂度： O(N * k * log(N * k))。
  所有元素的大小为n * k，因此时间复杂度为O(N * k * log(N * k))
• 空间复杂度： O(N * k)。
  要存储输出数组，需要O(N * k)空间。

高效的解决方案：
方法：一旦开始将k个数组视为合并排序算法的中间状态，这个想法就会变得清晰。
由于已经排序了k个数组，因此请合并k个数组。创建一个递归函数，该函数将采用k个数组并将其分为两部分，然后每半部分递归地调用该函数。基本情况是k的值小于3。
请参阅本文以在O(n)时间内合并两个数组。
算法：

1. 初始化大小为N * k的输出数组。
2. 调用函数除法。让$l$和$r$表示要合并的数组范围，因此在0到k-1之间变化。
3. 在每一步中，我们递归调用范围的左半部分和右半部分，以便将它们排序并存储在输出数组中。
4. 之后，我们合并左半部分和右半部分。对于合并，我们需要确定输出数组中左半部分和右半部分的索引范围。我们可以很容易地发现这一点。
   1. 左部分将从输出数组的索引l * n开始。
   2. 同样，右边部分将从输出数组的索引((l + r)/ 2 +1)* n开始。

C++

// C++ program to merge K
// sorted arrays
 
#include 
#define n 4
 
using namespace std;
 
// Function to perform merge operation
void merge(int l, int r, int* output)
{
    // to store the starting point
    // of left and right array
    int l_in = l * n, r_in
                      = ((l + r) / 2 + 1) * n;
 
    // To store the size of left and
    // right array
    int l_c = ((l + r) / 2 - l + 1) * n;
    int r_c = (r - (l + r) / 2) * n;
 
    // array to temporarily store left
    // and right array
    int l_arr[l_c], r_arr[r_c];
 
    // storing data in left array
    for (int i = 0; i < l_c; i++)
        l_arr[i] = output[l_in + i];
 
    // storing data in right array
    for (int i = 0; i < r_c; i++)
        r_arr[i] = output[r_in + i];
 
    // to store the current index of
    // temporary left and right array
    int l_curr = 0, r_curr = 0;
 
    // to store the current index
    // for output array
    int in = l_in;
 
    // two pointer merge for
    // two sorted arrays
    while (
        l_curr + r_curr < l_c + r_c) {
        if (
            r_curr == r_c
            || (l_curr != l_c
                && l_arr[l_curr] < r_arr[r_curr]))
            output[in]
                = l_arr[l_curr],
                l_curr++, in++;
        else
            output[in]
                = r_arr[r_curr],
                r_curr++, in++;
    }
}
 
// Code to drive merge-sort and
// create recursion tree
void divide(int l, int r, int* output,
            int arr[][n])
{
    if (l == r) {
 
        /* base step to initialize the output
           array before performing merge
           operation */
        for (int i = 0; i < n; i++)
            output[l * n + i] = arr[l][i];
 
        return;
    }
 
    // To sort left half
    divide(l, (l + r) / 2,
           output, arr);
 
    // To sort right half
    divide((l + r) / 2 + 1, r,
           output, arr);
 
    // Merge the left and right half
    merge(l, r, output);
}
 
// Driver Function
int main()
{
    // input 2D-array
    int arr[][n] = { { 5, 7, 15, 18 },
                     { 1, 8, 9, 17 },
                     { 1, 4, 7, 7 } };
 
    // Number of arrays
    int k = sizeof(arr) / sizeof(arr[0]);
 
    // Output array
    int* output = new int[n * k];
 
    divide(0, k - 1, output, arr);
 
    // Print merged array
    for (int i = 0; i < n * k; i++)
        cout << output[i] << " ";
 
    return 0;
}

Java

// Java program to merge
// K sorted arrays
import java.util.*;
 
class GFG {
 
    static int n = 4;
 
    // Function to perform
    // merge operation
    static void merge(
        int l, int r, int[] output)
    {
        // To store the starting point
        // of left and right array
        int l_in = l * n, r_in
                          = ((l + r) / 2 + 1) * n;
 
        // to store the size of left and
        // right array
        int l_c = ((l + r) / 2 - l + 1) * n;
        int r_c = (r - (l + r) / 2) * n;
 
        // array to temporarily store left
        // and right array
        int l_arr[] = new int[l_c],
            r_arr[] = new int[r_c];
 
        // storing data in left array
        for (int i = 0; i < l_c; i++)
            l_arr[i] = output[l_in + i];
 
        // storing data in right array
        for (int i = 0; i < r_c; i++)
            r_arr[i] = output[r_in + i];
 
        // to store the current index of
        // temporary left and right array
        int l_curr = 0, r_curr = 0;
 
        // to store the current index
        // for output array
        int in = l_in;
 
        // two pointer merge for two sorted arrays
        while (l_curr + r_curr < l_c + r_c) {
            if (
                r_curr == r_c
                || (l_curr != l_c
                    && l_arr[l_curr] < r_arr[r_curr])) {
                output[in] = l_arr[l_curr];
                l_curr++;
                in++;
            }
            else {
                output[in] = r_arr[r_curr];
                r_curr++;
                in++;
            }
        }
    }
 
    // Code to drive merge-sort and
    // create recursion tree
    static void divide(int l, int r, int[] output,
                       int arr[][])
    {
        if (l == r) {
 
            /* base step to initialize the output
        array before performing merge
        operation */
            for (int i = 0; i < n; i++)
                output[l * n + i] = arr[l][i];
 
            return;
        }
 
        // to sort left half
        divide(l, (l + r) / 2, output, arr);
 
        // to sort right half
        divide((l + r) / 2 + 1, r, output, arr);
 
        // merge the left and right half
        merge(l, r, output);
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        // input 2D-array
        int arr[][] = { { 5, 7, 15, 18 },
                        { 1, 8, 9, 17 },
                        { 1, 4, 7, 7 } };
 
        // Number of arrays
        int k = arr.length;
 
        // Output array
        int[] output = new int[n * k];
 
        divide(0, k - 1, output, arr);
 
        // Print merged array
        for (int i = 0; i < n * k; i++)
            System.out.print(output[i] + " ");
    }
}
 
/* This code contributed by PrinciRaj1992 */

Python3

# Python3 program to merge K sorted arrays
n = 4 ;
 
# Function to perform merge operation
def merge(l, r, output) :
     
    # to store the starting point of
    # left and right array
    l_in = l * n ;
    r_in = ((l + r) // 2 + 1) * n;
 
    # to store the size of left and
    # right array
    l_c = ((l + r) // 2 - l + 1) * n;
    r_c = (r - (l + r) // 2) * n;
 
    # array to temporarily store left
    # and right array
    l_arr = [0] * l_c; r_arr = [0] * r_c;
 
    # storing data in left array
    for i in range(l_c) :
        l_arr[i] = output[l_in + i];
 
    # storing data in right array
    for i in range(r_c) :
        r_arr[i] = output[r_in + i];
 
    # to store the current index of
    # temporary left and right array
    l_curr = 0 ; r_curr = 0;
 
    # to store the current index
    # for output array
    in1 = l_in;
 
    # two pointer merge for two sorted arrays
    while (l_curr + r_curr < l_c + r_c) :
        if (r_curr == r_c or (l_curr != l_c and
            l_arr[l_curr] < r_arr[r_curr])) :
            output[in1] = l_arr[l_curr];
            l_curr += 1; in1 += 1;
        else :
            output[in1] = r_arr[r_curr];
            r_curr += 1; in1 += 1;
 
# Code to drive merge-sort and
# create recursion tree
def divide(l, r, output, arr) :
     
    if (l == r) :
 
        # base step to initialize the output
        # array before performing merge
        # operation
        for i in range(n) :
            output[l * n + i] = arr[l][i];
 
        return;
     
    # to sort left half
    divide(l, (l + r) // 2, output, arr);
 
    # to sort right half
    divide((l + r) // 2 + 1, r, output, arr);
 
    # merge the left and right half
    merge(l, r, output);
 
# Driver code
if __name__ == "__main__" :
 
    # input 2D-array
    arr = [[ 5, 7, 15, 18 ],
           [ 1, 8, 9, 17 ],
           [ 1, 4, 7, 7 ]];
     
    # Number of arrays
    k = len(arr);
     
    # Output array
    output = [0] * (n * k);
     
    divide(0, k - 1, output, arr);
     
    # Print merged array
    for i in range(n * k) :
        print(output[i], end = " ");
 
# This code is contributed by Ryuga

C＃

// C# program to merge K sorted arrays
using System;
 
class GFG {
 
    static int n = 4;
 
    // Function to perform merge operation
    static void merge(int l, int r, int[] output)
    {
        // to store the starting point of left
        // and right array
        int l_in = l * n, r_in = ((l + r) / 2 + 1) * n;
 
        // to store the size of left and
        // right array
        int l_c = ((l + r) / 2 - l + 1) * n;
        int r_c = (r - (l + r) / 2) * n;
 
        // array to temporarily store left
        // and right array
        int[] l_arr = new int[l_c];
        int[] r_arr = new int[r_c];
 
        // storing data in left array
        for (int i = 0; i < l_c; i++)
            l_arr[i] = output[l_in + i];
 
        // storing data in right array
        for (int i = 0; i < r_c; i++)
            r_arr[i] = output[r_in + i];
 
        // to store the current index of
        // temporary left and right array
        int l_curr = 0, r_curr = 0;
 
        // to store the current index
        // for output array
        int index = l_in;
 
        // two pointer merge for two sorted arrays
        while (l_curr + r_curr < l_c + r_c) {
            if (r_curr == r_c || (l_curr != l_c && l_arr[l_curr] < r_arr[r_curr])) {
                output[index] = l_arr[l_curr];
                l_curr++;
                index++;
            }
            else {
                output[index] = r_arr[r_curr];
                r_curr++;
                index++;
            }
        }
    }
 
    // Code to drive merge-sort and
    // create recursion tree
    static void divide(int l, int r, int[] output,
                       int[, ] arr)
    {
        if (l == r) {
 
            /* base step to initialize the output
        array before performing merge
        operation */
            for (int i = 0; i < n; i++)
                output[l * n + i] = arr[l, i];
 
            return;
        }
 
        // to sort left half
        divide(l, (l + r) / 2, output, arr);
 
        // to sort right half
        divide((l + r) / 2 + 1, r, output, arr);
 
        // merge the left and right half
        merge(l, r, output);
    }
 
    // Driver Code
    public static void Main(String[] args)
    {
        // input 2D-array
        int[, ] arr = { { 5, 7, 15, 18 },
                        { 1, 8, 9, 17 },
                        { 1, 4, 7, 7 } };
 
        // Number of arrays
        int k = arr.GetLength(0);
 
        // Output array
        int[] output = new int[n * k];
 
        divide(0, k - 1, output, arr);
 
        // Print merged array
        for (int i = 0; i < n * k; i++)
            Console.Write(output[i] + " ");
    }
}
 
// This code has been contributed by 29AjayKumar

输出：

1 1 4 5 7 7 7 8 9 15 17 18

复杂度分析：

• 时间复杂度： O(N * k * log(k))。
  在每个级别中，遍历一次大小为N * k的数组，级别数为log(k)。
• 空间复杂度： O(N * k)。
  要存储输出数组，需要O(N * k)空间。

我们还可以通过使用min-heap解决此问题。