用于合并 3 个有序数组的 Python3 程序

给定 3 个按升序排序的数组（A、B、C），我们需要将它们按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）
我们在 Merging 2 Sorted arrays 中讨论过。所以我们可以先合并两个数组，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，这确实是该问题可以实现的最佳时间复杂度。
空间复杂性：由于我们一次合并两个数组，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，会忽略保存 3 个数组的结果所需的空间。

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)

下面给出了实现

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）
方法1的空间复杂度可以提高，我们将三个数组合并在一起。

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)，因为我们处理了三个数组中的每个元素一次。我们只需要一个数组来存储合并的结果，所以忽略这个数组，空间复杂度是 O(1)。

该算法的实现如下：

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))

输出

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

注意：虽然实现合并两个或三个数组的直接过程相对容易，但如果我们要合并 4 个或更多数组，该过程会变得很麻烦。在这种情况下，我们应该遵循 Merge K Sorted Arrays 中显示的过程。

有关更多详细信息，请参阅有关 Merge 3 Sorted Arrays 的完整文章！