在不使用额外空间的情况下，将数组 {a1, a2, .. an, b1, b2, .. bn} 随机播放为 {a1, b1, a2, b2, a3, b3, ......, an, bn}
给定一个由 2n 个元素组成的数组，格式如下 { a1, a2, a3, a4, ....., an, b1, b2, b3, b4, ...., bn }。任务是在不使用额外空间的情况下将数组随机排列为 {a1, b1, a2, b2, a3, b3, ......, an, bn }。
例子 ：
Input : arr[] = { 1, 2, 9, 15 }
Output : 1 9 2 15
Input : arr[] = { 1, 2, 3, 4, 5, 6 }
Output : 1 4 2 5 3 6
我们在下面的帖子中讨论了这个问题的 O(n 2 ) 和 O(n Log n) 解决方案。
以 {a1, b1, a2, b2, a3, b3, ……, an, bn} 格式随机播放 2n 个整数，无需使用额外空间
这里讨论了一种在线性时间内起作用的新解决方案。我们知道第一个和最后一个数字不会从它们的位置移动。并且，我们跟踪从中选择任何数字的索引以及目标索引的位置。我们知道，如果我们选择 ai，它必须去索引 2 * i – 1，如果是 bi，它必须去 2 * i。我们可以根据选择索引检查我们从哪里选择了某个数字，如果它大于或小于 n。
假设 n = 数组长度的一半，我们将不得不这样做 2 * n – 2 次。
我们得到两种情况，当 n 是偶数和奇数时，因此我们适当地初始化 start 变量。
注意：为简单起见，索引在数组中被视为 1。
C++
// C++ program to shuffle an array in O(n) time
// and O(1) extra space
#include
using namespace std;
// Shuffles an array of size 2n. Indexes
// are considered starting from 1.
void shufleArray(int a[], int n)
{
n = n / 2;
for (int start = n + 1, j = n + 1, done = 0, i;
done < 2 * n - 2; done++) {
if (start == j) {
start--;
j--;
}
i = j > n ? j - n : j;
j = j > n ? 2 * i : 2 * i - 1;
swap(a[start], a[j]);
}
}
// Driven Program
int main()
{
// The first element is bogus. We have used
// one based indexing for simplicity.
int a[] = { -1, 1, 3, 5, 7, 2, 4, 6, 8 };
int n = sizeof(a) / sizeof(a[0]);
shufleArray(a, n);
for (int i = 1; i < n; i++)
cout << a[i] << " ";
return 0;
}
Java
// Java program to shuffle an
// array in O(n) time and O(1)
// extra space
import java.io.*;
public class GFG {
// Shuffles an array of size 2n.
// Indexes are considered starting
// from 1.
static void shufleArray(int[] a, int n)
{
int temp;
n = n / 2;
for (int start = n + 1, j = n + 1, done = 0, i;
done < 2 * n - 2; done++) {
if (start == j) {
start--;
j--;
}
i = j > n ? j - n : j;
j = j > n ? 2 * i : 2 * i - 1;
temp = a[start];
a[start] = a[j];
a[j] = temp;
}
}
// Driver code
static public void main(String[] args)
{
// The first element is bogus. We have used
// one based indexing for simplicity.
int[] a = { -1, 1, 3, 5, 7, 2, 4, 6, 8 };
int n = a.length;
shufleArray(a, n);
for (int i = 1; i < n; i++)
System.out.print(a[i] + " ");
}
}
// This Code is contributed by vt_m.
Python3
# Python 3 program to shuffle an array
# in O(n) time and O(1) extra space
# Shuffles an array of size 2n. Indexes
# are considered starting from 1.
def shufleArray(a, n):
n = n // 2
start = n + 1
j = n + 1
for done in range( 2 * n - 2) :
if (start == j) :
start -= 1
j -= 1
i = j - n if j > n else j
j = 2 * i if j > n else 2 * i - 1
a[start], a[j] = a[j], a[start]
# Driver Code
if __name__ == "__main__":
# The first element is bogus. We have used
# one based indexing for simplicity.
a = [ -1, 1, 3, 5, 7, 2, 4, 6, 8 ]
n = len(a)
shufleArray(a, n)
for i in range(1, n):
print(a[i], end = " ")
# This code is contributed
# by ChitraNayal
C＃
// C# program to shuffle an
// array in O(n) time and O(1)
// extra space
using System;
public class GFG {
// Shuffles an array of size 2n.
// Indexes are considered starting
// from 1.
static void shufleArray(int[] a, int n)
{
int temp;
n = n / 2;
for (int start = n + 1, j = n + 1, done = 0, i;
done < 2 * n - 2; done++) {
if (start == j) {
start--;
j--;
}
i = j > n ? j - n : j;
j = j > n ? 2 * i : 2 * i - 1;
temp = a[start];
a[start] = a[j];
a[j] = temp;
}
}
// Driven code
static public void Main(String[] args)
{
// The first element is bogus. We have used
// one based indexing for simplicity.
int[] a = { -1, 1, 3, 5, 7, 2, 4, 6, 8 };
int n = a.Length;
shufleArray(a, n);
for (int i = 1; i < n; i++)
Console.Write(a[i] + " ");
}
}
// This Code is contributed by vt_m.
PHP
$k ? $j - $k : $j;
$j = $j > $k ? 2 * $i : 2 * $i - 1;
$temp = $a[$start];
$a[$start] = $a[$j];
$a[$j] = $temp;
}
for ($i = 1; $i < $n; $i++)
echo $a[$i] . " ";
}
// Driver code
// The first element is bogus.
// We have used one based
// indexing for simplicity.
$a = array(-1, 1, 3, 5, 7, 2, 4, 6, 8);
$n = count($a);
shufleArray($a, $n);
// This code is contributed by Sam007
?>