合并3个排序的数组

给定3个以升序排列的数组(A，B，C)，我们需要将它们以升序合并在一起并输出数组D。

例子：

Input : A = [1, 2, 3, 4, 5] 
        B = [2, 3, 4]
        C = [4, 5, 6, 7]
Output : D = [1, 2, 2, 3, 3, 4, 4, 4, 5, 5, 6, 7]

Input : A = [1, 2, 3, 5]
        B = [6, 7, 8, 9 ]
        C = [10, 11, 12]
Output: D = [1, 2, 3, 5, 6, 7, 8, 9. 10, 11, 12]

方法1(一次两个数组)
我们已经在合并2排序数组上进行了讨论。因此，我们可以首先合并两个数组，然后将结果与第三个数组合并。合并两个数组O(m + n)的时间复杂度。因此，对于合并第三个数组，时间复杂度将变为O(m + n + o)。请注意，这确实是可以解决此问题的最佳时间复杂度。
空间复杂度：由于我们一次合并两个数组，因此我们需要另一个数组来存储第一次合并的结果。这将空间复杂度提高到O(m + n)。请注意，在计算复杂度时，保留3个数组的结果所需的空间将被忽略。

算法

function merge(A, B)
    Let m and n be the sizes of A and B
    Let D be the array to store result
   
    // Merge by taking smaller element from A and B
    while i < m and j < n
        if A[i] <= B[j]
            Add A[i] to D and increment i by 1
        else Add B[j] to D and increment j by 1

    // If array A has exhausted, put elements from B
    while j < n
        Add B[j] to D and increment j by 1
   
    // If array B has exhausted, put elements from A
    while i < n
        Add A[j] to D and increment i by 1
   
    Return D

function merge_three(A, B, C)
    T = merge(A, B)
    return merge(T, C)

具体实现如下

C++

// C++ program to merge three sorted arrays
// by merging two at a time.
#include 
#include 
using namespace std;
 
using Vector = vector;
 
void printVector(const Vector& a)
{
    cout << "[";
    for (auto e : a)
        cout << e << " ";
    cout << "]" << endl;
}
 
Vector mergeTwo(Vector& A, Vector& B)
{
    // Get sizes of vectors
    int m = A.size();
    int n = B.size();
 
    // Vector for storing Result
    Vector D;
    D.reserve(m + n);
 
    int i = 0, j = 0;
    while (i < m && j < n) {
 
        if (A[i] <= B[j])
            D.push_back(A[i++]);
        else
            D.push_back(B[j++]);
    }
 
    // B has exhausted
    while (i < m)
        D.push_back(A[i++]);
 
    // A has exhausted
    while (j < n)
        D.push_back(B[j++]);
 
    return D;
}
 
// Driver Code
int main()
{
    Vector A = { 1, 2, 3, 5 };
    Vector B = { 6, 7, 8, 9 };
    Vector C = { 10, 11, 12 };
 
    // First Merge A and B
    Vector T = mergeTwo(A, B);
 
    // Print Result after merging T with C
    printVector(mergeTwo(T, C));
    return 0;
}

Java

import java.util.*;
// Java program to merge three sorted arrays
// by merging two at a time.
class GFG {
 
    static ArrayList mergeTwo(List A,
                                       List B)
    {
        // Get sizes of vectors
        int m = A.size();
        int n = B.size();
 
        // ArrayList for storing Result
        ArrayList D = new ArrayList(m + n);
 
        int i = 0, j = 0;
        while (i < m && j < n) {
 
            if (A.get(i) <= B.get(j))
                D.add(A.get(i++));
            else
                D.add(B.get(i++));
        }
 
        // B has exhausted
        while (i < m)
            D.add(A.get(i++));
 
        // A has exhausted
        while (j < n)
            D.add(B.get(j++));
 
        return D;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        Integer[] a = { 1, 2, 3, 5 };
        Integer[] b = { 6, 7, 8, 9 };
        Integer[] c = { 10, 11, 12 };
        List A = Arrays.asList(a);
        List B = Arrays.asList(b);
        List C = Arrays.asList(c);
 
        // First Merge A and B
        ArrayList T = mergeTwo(A, B);
 
        // Print Result after merging T with C
        System.out.println(mergeTwo(T, C));
    }
}
 
/* This code contributed by PrinciRaj1992 */

Python

# Python program to merge three sorted arrays
# by merging two at a time.
 
def merge_two(a, b):
    (m, n) = (len(a), len(b))
    i = j = 0
 
    # Destination Array
    d = []
 
    # Merge from a and b together
    while i < m and j < n:
        if a[i] <= b[j]:
            d.append(a[i])
            i += 1
        else:
            d.append(b[j])
            j += 1
 
    # Merge from a if b has run out
    while i < m:
        d.append(a[i])
        i += 1
 
    # Merge from b if a has run out
    while j < n:
        d.append(b[j])
        j += 1
 
    return d
 
def merge(a, b, c):
    t = merge_two(a, b)
    return merge_two(t, c)
 
if __name__ == "__main__":
    A = [1, 2, 3, 5]
    B = [6, 7, 8, 9]
    C = [10, 11, 12]
    print(merge(A, B, C))

输出：

[1, 2, 3, 5, 6, 7, 8, 9, 10, 11, 12]

方法2(一次三个数组)
通过将三个数组合并在一起，可以提高方法1的空间复杂度。

function merge-three(A, B, C)
    Let m, n, o be size of A, B, and C
    Let D be the array to store the result
    
    // Merge three arrays at the same time
    while i < m and j < n and k < o
        Get minimum of A[i], B[j], C[i]
        if the minimum is from A, add it to 
           D and advance i
        else if the minimum is from B add it 
                to D and advance j
        else if the minimum is from C add it 
                to D and advance k
    
   // After above step at least 1 array has 
   // exhausted. Only C has exhausted
   while i < m and j < n
       put minimum of A[i] and B[j] into D
       Advance i if minimum is from A else advance j 
   
   // Only B has exhausted
   while i < m and k < o
       Put minimum of A[i] and C[k] into D
       Advance i if minimum is from A else advance k
 
   // Only A has exhausted
   while j < n and k < o
       Put minimum of B[j] and C[k] into D
       Advance j if minimum is from B else advance k

   // After above steps at least 2 arrays have 
   // exhausted
   if A and B have exhausted take elements from C
   if B and C have exhausted take elements from A
   if A and C have exhausted take elements from B
   
   return D

复杂度：时间复杂度为O(m + n + o)，因为我们一次处理了三个数组中的每个元素。我们只需要一个数组来存储合并的结果，因此忽略该数组，空间复杂度为O(1)。

该算法的实现如下：

C++

// C++ program to merger three sorted arrays
// by merging three simultaneously.
#include 
#include 
using namespace std;
 
using Vector = vector;
 
void printVector(const Vector& a)
{
    cout << "[";
    for (auto e : a) {
        cout << e << " ";
    }
    cout << "]" << endl;
}
 
Vector mergeThree(Vector& A, Vector& B,
                  Vector& C)
{
    int m, n, o, i, j, k;
    // Get Sizes of three vectors
    m = A.size();
    n = B.size();
    o = C.size();
 
    // Vector for storing output
    Vector D;
    D.reserve(m + n + o);
 
    i = j = k = 0;
 
    while (i < m && j < n && k < o) {
 
        // Get minimum of a, b, c
        int m = min(min(A[i], B[j]), C[k]);
 
        // Put m in D
        D.push_back(m);
 
        // Increment i, j, k
        if (m == A[i])
            i++;
        else if (m == B[j])
            j++;
        else
            k++;
    }
 
    // C has exhausted
    while (i < m && j < n) {
        if (A[i] <= B[j]) {
            D.push_back(A[i]);
            i++;
        }
        else {
            D.push_back(B[j]);
            j++;
        }
    }
 
    // B has exhausted
    while (i < m && k < o) {
        if (A[i] <= C[k]) {
            D.push_back(A[i]);
            i++;
        }
        else {
            D.push_back(C[k]);
            k++;
        }
    }
 
    // A has exhausted
    while (j < n && k < o) {
        if (B[j] <= C[k]) {
            D.push_back(B[j]);
            j++;
        }
        else {
            D.push_back(C[k]);
            k++;
        }
    }
 
    // A and B have exhausted
    while (k < o)
        D.push_back(C[k++]);
 
    // B and C have exhausted
    while (i < m)
        D.push_back(A[i++]);
 
    // A and C have exhausted
    while (j < n)
        D.push_back(B[j++]);
 
    return D;
}
 
// Driver Code
int main()
{
    Vector A = { 1, 2, 41, 52, 84 };
    Vector B = { 1, 2, 41, 52, 67 };
    Vector C = { 1, 2, 41, 52, 67, 85 };
 
    // Print Result
    printVector(mergeThree(A, B, C));
    return 0;
}

Python

# Python program to merge three sorted arrays
# simultaneously.
 
def merge_three(a, b, c):
    (m, n, o) = (len(a), len(b), len(c))
    i = j = k = 0
 
    # Destination array
    d = []
 
    while i < m and j < n and k < o:
 
        # Get Minimum element
        m = min(a[i], b[j], c[k])
 
        # Add m to D
        d.append(m)
 
        # Increment the source pointer which
        # gives m
        if a[i] == m:
            i += 1
        elif b[j] == m:
            j += 1
        elif c[k] == m:
            k += 1
 
    # Merge a and b in c has exhausted
    while i < m and j < n:
        if a[i] <= b[k]:
            d.append(a[i])
            i += 1
        else:
            d.append(b[j])
            j += 1
 
    # Merge b and c if a has exhausted
    while j < n and k < o:
        if b[j] <= c[k]:
            d.append(b[j])
            j += 1
        else:
            d.append(c[k])
            k += 1
 
    # Merge a and c if b has exhausted
    while i < m and k < o:
        if a[i] <= c[k]:
            d.append(a[i])
            i += 1
        else:
            d.append(c[k])
            k += 1
 
    # Take elements from a if b and c
    # have exhausted
    while i < m:
        d.append(a[i])
        i += 1
 
    # Take elements from b if a and c
    # have exhausted
    while j < n:
        d.append(b[j])
        j += 1
 
    # Take elements from c if a and
    # b have exhausted
    while k < o:
        d.append(c[k])
        k += 1
 
    return d
 
if __name__ == "__main__":
    a = [1, 2, 41, 52, 84]
    b = [1, 2, 41, 52, 67]
    c = [1, 2, 41, 52, 67, 85]
 
    print(merge_three(a, b, c))

Java

import java.util.*;
import java.io.*;
import java.lang.*;
class Sorting {
    public static void main(String[] args)
    {
        int A[] = { 1, 2, 41, 52, 84 };
        int B[] = { 1, 2, 41, 52, 67 };
        int C[] = { 1, 2, 41, 52, 67, 85 };
 
        // call the function to sort and print the sorted numbers
        merge3sorted(A, B, C);
    }
 
    // Function to merge three sorted arrays
    // A[], B[], C[]: input arrays
    static void merge3sorted(int A[], int B[], int C[])
    {
        // creating an empty list to store sorted numbers
        ArrayList list = new ArrayList();
        int i = 0, j = 0, k = 0;
 
        // using merge concept and trying to find
        // smallest of three while all three arrays
        // contains at least one element
        while (i < A.length && j < B.length && k < C.length) {
            int a = A[i];
            int b = B[j];
            int c = C[k];
            if (a <= b && a <= c) {
                list.add(a);
                i++;
            }
            else if (b <= a && b <= c) {
                list.add(b);
                j++;
            }
            else {
                list.add(c);
                k++;
            }
        }
        // next three while loop is to sort two
        // of arrays if one of the three gets exhausted
        while (i < A.length && j < B.length) {
            if (A[i] < B[j]) {
                list.add(A[i]);
                i++;
            }
            else {
                list.add(B[j]);
                j++;
            }
        }
        while (j < B.length && k < C.length) {
            if (B[j] < C[k]) {
                list.add(B[j]);
                j++;
            }
            else {
                list.add(C[k]);
                k++;
            }
        }
        while (i < A.length && k < C.length) {
            if (A[i] < C[k]) {
                list.add(A[i]);
                i++;
            }
            else {
                list.add(C[k]);
                k++;
            }
        }
 
        // if one of the array are left then
        // simply appending them as there will
        // be only largest element left
        while (i < A.length) {
            list.add(A[i]);
            i++;
        }
        while (j < B.length) {
            list.add(B[j]);
            j++;
        }
        while (k < C.length) {
            list.add(C[k]);
            k++;
        }
        // finally print the list
        for (Integer x : list)
            System.out.print(x + " ");
    } // merge3sorted closing braces
}

输出

[1 1 1 2 2 2 41 41 41 52 52 52 67 67 84 85 ]

注意：虽然实现直接过程合并两个或三个数组相对容易，但是如果我们要合并4个或更多数组，则过程变得很麻烦。在这种情况下，我们应该遵循合并K排序数组中显示的过程。