用于合并 3 个排序数组的 Javascript 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
Javascript
Javascript
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:
Javascript
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!