给定两个排序数组arr1[]和arr2[] ，任务是将它们在O(Nlog(N) + Mlog(M))时间内以O(1)额外空间合并成一个排序数组，其中N是第一个数组arr1[]和M是第二个数组arr2[]的大小。
例子：
Input: arr1[] = {1, 5, 9, 10, 15, 20}, arr2[] = {2, 3, 8, 13}
Output: 1 2 3 5 8 9 10 13 15 20
Input: arr1[] = {4, 9, 15, 20}, arr2[] = {2, 6, 7, 13}
Output: 2 4 6 7 9 13 15 20
方法：本文已经讨论了一种方法。在本文中，将讨论一种甚至有效的方法。
- 通过比较一个数组的最高元素与第二个数组的最低元素来形成两个双调数组，这样两个数组都只包含两个数组排序后将在那里的元素。
- 现在，分别对两个数组进行排序。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Reducing the gap by a factor of 2
int nextGap(int gap)
{
if (gap <= 1)
return 0;
return (gap / 2) + (gap % 2);
}
// Function to merge two sorted
// arrays with O(1) extra space
int mergeTwoSortedArray(int* arr1,
int* arr2,
int n, int m)
{
int x = min(n, m);
// Form both arrays to be bitonic
for (int i = 0; i < x; i++) {
if (arr1[n - i - 1] > arr2[i])
swap(arr1[n - i - 1], arr2[i]);
}
// Now both the arrays conatin the numbers
// which should be there in the result
// Sort the array indiviually by inplace algo
for (int gap = nextGap(n); gap > 0;
gap = nextGap(gap)) {
// Comparing elements in the first array
for (int i = 0; i + gap < n; i++)
if (arr1[i] > arr1[i + gap])
swap(arr1[i], arr1[i + gap]);
}
// Sort the second array
for (int gap = nextGap(m); gap > 0;
gap = nextGap(gap)) {
// Comparing elements in the second array
for (int i = 0; i + gap < m; i++)
if (arr2[i] > arr2[i + gap])
swap(arr2[i], arr2[i + gap]);
}
for (int i = 0; i < n; i++)
cout << arr1[i] << " ";
for (int j = 0; j < m; j++)
cout << arr2[j] << " ";
}
// Driver code
int main()
{
int arr1[] = { 1, 5, 9, 10, 15, 20 };
int n = sizeof(arr1) / sizeof(int);
int arr2[] = { 2, 3, 8, 13 };
int m = sizeof(arr2) / sizeof(int);
mergeTwoSortedArray(arr1, arr2, n, m);
return 0;
} Java
// Java implementation of the approach
class GFG
{
// Reducing the gap by a factor of 2
static int nextGap(int gap)
{
if (gap <= 1)
return 0;
return (int)((gap / 2) + (gap % 2));
}
// Function to merge two sorted
// arrays with O(1) extra space
static void mergeTwoSortedArray(int []arr1,
int []arr2,
int n, int m)
{
int x = Math.min(n, m);
// Form both arrays to be bitonic
for (int i = 0; i < x; i++)
{
if (arr1[n - i - 1] > arr2[i])
{
// swap(arr1[n - i - 1], arr2[i]);
int temp = arr1[n - i - 1];
arr1[n - i - 1] = arr2[i];
arr2[i] = temp;
}
}
// Now both the arrays conatin the numbers
// which should be there in the result
// Sort the array indiviually by inplace algo
for (int gap = nextGap(n); gap > 0;
gap = nextGap(gap))
{
// Comparing elements in the first array
for (int i = 0; i + gap < n; i++)
if (arr1[i] > arr1[i + gap])
{
// swap(arr1[i], arr1[i + gap]);
int temp = arr1[i];
arr1[i] = arr1[i + gap];
arr1[i + gap] = temp;
}
}
// Sort the second array
for (int gap = nextGap(m); gap > 0;
gap = nextGap(gap))
{
// Comparing elements in the second array
for (int i = 0; i + gap < m; i++)
if (arr2[i] > arr2[i + gap])
{
// swap(arr2[i], arr2[i + gap]);
int temp = arr2[i];
arr2[i] = arr2[i + gap];
arr2[i + gap] = temp;
}
}
for (int i = 0; i < n; i++)
System.out.print(arr1[i] + " ");
for (int j = 0; j < m; j++)
System.out.print(arr2[j] + " ");
}
// Driver code
public static void main (String[] args)
{
int arr1[] = { 1, 5, 9, 10, 15, 20 };
int n = arr1.length;
int arr2[] = { 2, 3, 8, 13 };
int m = arr2.length;
mergeTwoSortedArray(arr1, arr2, n, m);
}
}
// This code is contributed by AnkitRai01Python3
# Python3 implementation of the approach
# Reducing the gap by a factor of 2
def nextGap(gap) :
if (gap <= 1) :
return 0;
res = (gap // 2) + (gap % 2);
return res;
# Function to merge two sorted
# arrays with O(1) extra space
def mergeTwoSortedArray(arr1, arr2, n, m) :
x = min(n, m);
# Form both arrays to be bitonic
for i in range(x) :
if (arr1[n - i - 1] > arr2[i]) :
arr1[n - i - 1],arr2[i] = arr2[i], arr1[n- i - 1];
# Now both the arrays conatin the numbers
# which should be there in the result
# Sort the array indiviually by inplace algo
gap = nextGap(n);
while gap > 0 :
# Comparing elements in the first array
i = 0;
while i + gap < n :
if (arr1[i] > arr1[i + gap]) :
arr1[i], arr1[i + gap] = arr1[i + gap],arr1[i];
i += 1;
gap = nextGap(gap)
# Sort the second array
gap = nextGap(m);
while gap > 0 :
# Comparing elements in the second array
i = 0
while i + gap < m :
if (arr2[i] > arr2[i + gap]) :
arr2[i], arr2[i + gap] = arr2[i + gap], arr2[i];
i += 1;
gap = nextGap(gap)
for i in range(n) :
print(arr1[i], end = " ");
for j in range(m) :
print(arr2[j], end = " ");
# Driver code
if __name__ == "__main__" :
arr1 = [ 1, 5, 9, 10, 15, 20 ];
n = len(arr1);
arr2 = [ 2, 3, 8, 13 ];
m = len(arr2);
mergeTwoSortedArray(arr1, arr2, n, m);
# This code is contributed by AnkitRai01C#
// C# implementation of the approach
using System;
class GFG
{
// Reducing the gap by a factor of 2
static int nextGap(int gap)
{
if (gap <= 1)
return 0;
return (int)((gap / 2) + (gap % 2));
}
// Function to merge two sorted
// arrays with O(1) extra space
static void mergeTwoSortedArray(int []arr1,
int []arr2,
int n, int m)
{
int x = Math.Min(n, m);
// Form both arrays to be bitonic
for (int i = 0; i < x; i++)
{
if (arr1[n - i - 1] > arr2[i])
{
int temp = arr1[n - i - 1];
arr1[n - i - 1] = arr2[i];
arr2[i] = temp;
}
}
// Now both the arrays conatin the numbers
// which should be there in the result
// Sort the array indiviually by inplace algo
for (int gap = nextGap(n); gap > 0;
gap = nextGap(gap))
{
// Comparing elements in the first array
for (int i = 0; i + gap < n; i++)
if (arr1[i] > arr1[i + gap])
{
int temp = arr1[i];
arr1[i] = arr1[i + gap];
arr1[i + gap] = temp;
}
}
// Sort the second array
for (int gap = nextGap(m); gap > 0;
gap = nextGap(gap))
{
// Comparing elements in the second array
for (int i = 0; i + gap < m; i++)
if (arr2[i] > arr2[i + gap])
{
int temp = arr2[i];
arr2[i] = arr2[i + gap];
arr2[i + gap] = temp;
}
}
for (int i = 0; i < n; i++)
Console.Write(arr1[i] + " ");
for (int j = 0; j < m; j++)
Console.Write(arr2[j] + " ");
}
// Driver code
public static void Main()
{
int []arr1 = { 1, 5, 9, 10, 15, 20 };
int n = arr1.Length;
int []arr2 = { 2, 3, 8, 13 };
int m = arr2.Length;
mergeTwoSortedArray(arr1, arr2, n, m);
}
}
// This code is contributed by AnkitRai01输出：
1 2 3 5 8 9 10 13 15 20
时间复杂度： O(Nlog(N) + Mlog(M))
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