给定长度为n + 1的数组，其中包含元素1到n和一个空格，需要使用给定的swap(索引i，索引j)函数对数组进行排序，您只能在中间交换空格和数字。最后，把差距放在最后。
数组中的数字999将作为间隙或间隔。
例子：
Input : arr = {1, 5, 4, 999, 3, 2}
Output : arr = {1, 2, 3, 4, 5, 999}
We need to sort only by moving
Input : arr = {1, 5, 4, 3, 2, 8, 7, 999, 6}
Output : arr = {1, 2, 3, 4, 5, 6, 7, 8, 999}
我们遵循一种递归方法来解决此问题。因为我们只能用空格交换数字。首先我们找到空间的索引。如果索引是数组的开始，则通过与右边的每个数字交换将空间移到倒数第二个索引。
如果space既不是数组的开始，也不是array和element之前的最后一个元素，而不是空格旁边的元素，请执行以下操作。
Step 1: Swap space and element next to space
In case of {3, 999, 2} make it {3, 2, 999}
Step 2 : Swap space and greater element
eg-convert {3, 2, 999} to {999, 2, 3}
否则，将对index旁边的元素进行排序，并将其与上一个元素交换。再次调用sort函数，将数组的大小减小1，将空间索引减小1，因为我们每次都会得到一个排序的元素。
C++
// CPP program to sort an array by moving one
// space around.
#include
using namespace std;
// n is total number of elements.
// index is index of 999 or space.
// k is number of elements yet to be sorted.
void sortRec(int arr[], int index, int k, int n)
{
// print the sorted array when loop reaches
// the base case
if (k == 0) {
for (int i = 1; i < n; i++)
cout << arr[i] << " ";
cout << 999;
return;
}
// else if k>0 and space is at 0th index swap
// each number with space and store index at
// second last index
else if (k > 0 && index == 0) {
index = n - 2;
for (int i = 1; i <= index; i++) {
arr[i - 1] = arr[i];
}
arr[index] = 999;
}
// if space is neither start of array nor last
// element of array and element before it greater
// than/ the element next to space
if (index - 1 >= 0 && index + 1 < n &&
arr[index - 1] > arr[index + 1]) {
// first swap space and element next to space
// in case of {3, 999, 2} make it {3, 2, 999}
swap(arr[index], arr[index + 1]);
// than swap space and greater element
// convert {3, 2, 999} to {999, 2, 3}
swap(arr[index - 1], arr[index + 1]);
}
else
swap(arr[index], arr[index - 1]);
sortRec(arr, index - 1, k - 1, n);
}
// Wrapper over sortRec.
void sort(int arr[], int n)
{
// Find index of space (or 999)
int index = -1;
for (int i = 0; i < n; i++) {
if (arr[i] == 999) {
index = i;
break;
}
}
// Invalid input
if (index == -1)
return;
sortRec(arr, index, n, n);
}
// driver program
int main()
{
int arr[] = { 3, 2, 999, 1 };
int n = sizeof(arr) / sizeof(arr[0]);
sort(arr, n);
return 0;
} Java
// Java program to sort an array by moving one
// space around.
class GFG
{
// n is total number of elements.
// index is index of 999 or space.
// k is number of elements yet to be sorted.
static void sortRec(int arr[], int index, int k, int n)
{
// print the sorted array when loop reaches
// the base case
if (k == 0)
{
for (int i = 1; i < n; i++)
System.out.print(arr[i] + " ");
System.out.println(999);
return;
}
// else if k>0 and space is at 0th index swap
// each number with space and store index at
// second last index
else if (k > 0 && index == 0)
{
index = n - 2;
for (int i = 1; i <= index; i++)
{
arr[i - 1] = arr[i];
}
arr[index] = 999;
}
// if space is neither start of array nor last
// element of array and element before it greater
// than/ the element next to space
if (index - 1 >= 0 && index + 1 < n &&
arr[index - 1] > arr[index + 1])
{
// first swap space and element next to space
// in case of {3, 999, 2} make it {3, 2, 999}
swap(arr,index, index + 1);
// than swap space and greater element
// convert {3, 2, 999} to {999, 2, 3}
swap(arr,index - 1, index + 1);
}
else
swap(arr,index, index - 1);
sortRec(arr, index - 1, k - 1, n);
}
static int[] swap(int []arr, int i, int j)
{
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
return arr;
}
// Wrapper over sortRec.
static void sort(int arr[], int n)
{
// Find index of space (or 999)
int index = -1;
for (int i = 0; i < n; i++)
{
if (arr[i] == 999)
{
index = i;
break;
}
}
// Invalid input
if (index == -1)
return;
sortRec(arr, index, n, n);
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 3, 2, 999, 1 };
int n = arr.length;
sort(arr, n);
}
}
// This code contributed by Rajput-JiC#
// C# program to sort an array by moving one
// space around.
using System;
class GFG
{
// n is total number of elements.
// index is index of 999 or space.
// k is number of elements yet to be sorted.
static void sortRec(int []arr, int index, int k, int n)
{
// print the sorted array when loop reaches
// the base case
if (k == 0)
{
for (int i = 1; i < n; i++)
Console.Write(arr[i] + " ");
Console.WriteLine(999);
return;
}
// else if k>0 and space is at 0th index swap
// each number with space and store index at
// second last index
else if (k > 0 && index == 0)
{
index = n - 2;
for (int i = 1; i <= index; i++)
{
arr[i - 1] = arr[i];
}
arr[index] = 999;
}
// if space is neither start of array nor last
// element of array and element before it greater
// than/ the element next to space
if (index - 1 >= 0 && index + 1 < n &&
arr[index - 1] > arr[index + 1])
{
// first swap space and element next to space
// in case of {3, 999, 2} make it {3, 2, 999}
swap(arr,index, index + 1);
// than swap space and greater element
// convert {3, 2, 999} to {999, 2, 3}
swap(arr,index - 1, index + 1);
}
else
swap(arr,index, index - 1);
sortRec(arr, index - 1, k - 1, n);
}
static int[] swap(int []arr, int i, int j)
{
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
return arr;
}
// Wrapper over sortRec.
static void sort(int []arr, int n)
{
// Find index of space (or 999)
int index = -1;
for (int i = 0; i < n; i++)
{
if (arr[i] == 999)
{
index = i;
break;
}
}
// Invalid input
if (index == -1)
return;
sortRec(arr, index, n, n);
}
// Driver code
public static void Main(String[] args)
{
int []arr = { 3, 2, 999, 1 };
int n = arr.Length;
sort(arr, n);
}
}
// This code has been contributed by 29AjayKumar
输出：
1 2 3 999
时间复杂度-O(n 2 )