分离消极和积极的维持秩序和 O(1) 空间
在不使用额外空间的情况下分离数组中的负数和正数,并保持插入顺序和 O(n) 时间复杂度。
例子:
Input :9
12 11 -13 -5 6 -7 5 -3 -6
Output :-13 -5 -7 -3 -6 12 11 6 5
Input :5
11 -13 6 -7 5
Output :-13 -7 11 6 5
我们在下面的帖子中讨论了这个问题。
- ers-beginning-positive-end-constant-extra-space/”>在不保持顺序的情况下重新排列正数和负数。
- 用恒定的额外空间重新排列正数和负数
在这篇文章中,我们讨论了一种占用 O(1) 额外空间的新方法。我们首先计算总负数,然后将负数一个一个移动到正确的位置。
C++
// C++ program to move all negative numbers
// to beginning and positive numbers to end
// keeping order.
#include
using namespace std;
void segregate(int arr[], int n)
{
// Count negative numbers
int count_negative = 0;
for (int i = 0; i < n; i++)
if (arr[i] < 0)
count_negative++;
// Run a loop until all negative
// numbers are moved to the beginning
int i = 0, j = i + 1;
while (i != count_negative) {
// If number is negative, update
// position of next positive number.
if (arr[i] < 0) {
i++;
j = i + 1;
}
// If number is positive, move it to
// index j and increment j.
else if (arr[i] > 0 && j < n) {
swap(arr[i], arr[j]);
j++;
}
}
}
int main()
{
int count_negative = 0;
int arr[] = { -12, 11, -13, -5, 6, -7, 5, -3, -6 };
int n = sizeof(arr) / sizeof(arr[0]);
segregate(arr, n);
for (int i = 0; i < n; i++)
cout << arr[i] << " ";
}
Java
// Java program to move all
// negative numbers to beginning
// and positive numbers to end
// keeping order.
class GFG
{
static void segregate(int arr[],
int n)
{
// Count negative numbers
int count_negative = 0;
for (int i = 0; i < n; i++)
if (arr[i] < 0)
count_negative++;
// Run a loop until all
// negative numbers are
// moved to the beginning
int i = 0, j = i + 1;
while (i != count_negative)
{
// If number is negative,
// update position of next
// positive number.
if (arr[i] < 0)
{
i++;
j = i + 1;
}
// If number is positive, move
// it to index j and increment j.
else if (arr[i] > 0 && j < n)
{
int t = arr[i];
arr[i] = arr[j];
arr[j] = t;
j++;
}
}
}
// Driver code
public static void main(String[] args)
{
int count_negative = 0;
int arr[] = { -12, 11, -13, -5,
6, -7, 5, -3, -6 };
int n = arr.length;
segregate(arr, n);
for (int i = 0; i < n; i++)
System.out.print(arr[i] + " ");
}
}
// This code is contributed
// by ChitraNayal
C#
// C# program to move all
// negative numbers to beginning
// and positive numbers to end
// keeping order.
using System;
class GFG
{
static void segregate(int[] arr,
int n)
{
// Count negative numbers
int count_negative = 0,i;
for (i = 0; i < n; i++)
if (arr[i] < 0)
count_negative++;
// Run a loop until all
// negative numbers are
// moved to the beginning
i = 0;
int j = i + 1;
while (i != count_negative)
{
// If number is negative,
// update position of next
// positive number.
if (arr[i] < 0)
{
i++;
j = i + 1;
}
// If number is positive, move
// it to index j and increment j.
else if (arr[i] > 0 && j < n)
{
int t = arr[i];
arr[i] = arr[j];
arr[j] = t;
j++;
}
}
}
// Driver code
public static void Main()
{
int[] arr = { -12, 11, -13, -5,
6, -7, 5, -3, -6 };
int n = arr.Length;
segregate(arr, n);
for (int i = 0; i < n; i++)
Console.Write(arr[i] + " ");
}
}
// This code is contributed
// by ChitraNayal
Python 3
# Python 3 program to move all
# negative numbers to beginning
# and positive numbers to end
# keeping order.
def segregate(arr, n):
# Count negative numbers
count_negative = 0
for i in range(n):
if (arr[i] < 0):
count_negative += 1
# Run a loop until all
# negative numbers are
# moved to the beginning
i = 0
j = i + 1
while (i != count_negative):
# If number is negative,
# update position of next
# positive number.
if (arr[i] < 0) :
i += 1
j = i + 1
# If number is positive, move
# it to index j and increment j.
elif (arr[i] > 0 and j < n):
t = arr[i]
arr[i] = arr[j]
arr[j] = t
j += 1
# Driver Code
count_negative = 0
arr = [-12, 11, -13, -5,
6, -7, 5, -3, -6 ]
segregate(arr, 9)
for i in range(9):
print(arr[i] , end =" ")
# This code is contributed
# by ChitraNayal
PHP
0 && $j < $n)
{
$t = $arr[$i];
$arr[$i] = $arr[$j];
$arr[$j] = $t;
$j++;
}
}
}
// Driver Code
$count_negative = 0;
$arr = array(-12, 11, -13, -5,
6, -7, 5, -3, -6);
$n = sizeof($arr);
segregate($arr, $n);
for ($i = 0; $i < $n; $i++)
echo $arr[$i] ." ";
// This code is contributed
// by ChitraNayal
?>
Javascript
输出:
-12 -13 -5 -7 -3 -6 11 6 5