先决条件：循环排序
给定从1 到 N元素的数组arr[] ，任务是在O(N)时间内对给定数组进行排序。
例子：
Input: arr[] = { 2, 1, 5, 4, 3}
Output: 1 2 3 4 5
Explanation:
Since arr[0] = 2 is not at correct position, then swap arr[0] with arr[arr[0] – 1]
Now array becomes: arr[] = {1, 2, 5, 4, 3}
Now arr[2] = 5 is not at correct position, then swap arr[3] with arr[arr[3] – 1]
Now the array becomes: arr[] = {1, 2, 3, 4, 5}
Now the array is sorted.
Input: arr[] = {1, 2, 3, 4, 5, 6}
Output: 1 2 3 4 5 6
Explanation:
The array is already sorted.
方法：这个问题可以使用贪心方法来解决。以下是步骤：
- 遍历给定的数组arr[] 。
- 如果当前元素不在正确位置，即arr[i] 不等于 i+1 ，则将当前元素与具有正确位置的元素交换。
例如：
Let arr[] = {2, 1, 4, 5, 3}
Since, arr[0] = 2, which is not at it’s correct position 1.
Then swap this element with it’s correct position, i.e., swap(arr[0], arr[2-1])
Then array becomes: arr[] = {1, 2, 4, 5, 3}
- 如果当前元素位于正确位置，则检查下一个元素。
- 重复上述步骤，直到我们到达数组的末尾。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include "bits/stdc++.h"
using namespace std;
// Function to swap two a & b value
void swap(int* a, int* b)
{
int temp = *a;
*a = *b;
*b = temp;
}
// Function to print array element
void printArray(int arr[], int N)
{
// Traverse the array
for (int i = 0; i < N; i++) {
cout << arr[i] << ' ';
}
}
// Function to sort the array in O(N)
void sortArray(int arr[], int N)
{
// Traverse the array
for (int i = 0; i < N;) {
// If the current element is
// at correct position
if (arr[i] == i + 1) {
i++;
}
// Else swap the current element
// with it's correct position
else {
swap(&arr[i], &arr[arr[i] - 1]);
}
}
}
// Driver Code
int main()
{
int arr[] = { 2, 1, 5, 3, 4 };
int N = sizeof(arr) / sizeof(arr[0]);
// Function call to sort the array
sortArray(arr, N);
// Function call to print the array
printArray(arr, N);
return 0;
}Java
// Java program for the above approach
class Main{
// Function to print array element
public static void printArray(int arr[], int N)
{
// Traverse the array
for(int i = 0; i < N; i++)
{
System.out.print(arr[i] + " ");
}
}
// Function to sort the array in O(N)
public static void sortArray(int arr[], int N)
{
// Traverse the array
for(int i = 0; i < N;)
{
// If the current element is
// at correct position
if (arr[i] == i + 1)
{
i++;
}
// Else swap the current element
// with it's correct position
else
{
// Swap the value of
// arr[i] and arr[arr[i]-1]
int temp1 = arr[i];
int temp2 = arr[arr[i] - 1];
arr[i] = temp2;
arr[temp1 - 1] = temp1;
}
}
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 2, 1, 5, 3, 4 };
int N = arr.length;
// Function call to sort the array
sortArray(arr, N);
// Function call to print the array
printArray(arr, N);
}
}
// This code is contributed by divyeshrabadiya07Python3
# Python3 program for the above approach
# Function to print array element
def printArray(arr, N):
# Traverse the array
for i in range(N):
print(arr[i], end = ' ')
# Function to sort the array in O(N)
def sortArray(arr, N):
i = 0
# Traverse the array
while i < N:
# If the current element is
# at correct position
if arr[i] == i + 1:
i += 1
# Else swap the current element
# with it's correct position
else:
# Swap the value of
# arr[i] and arr[arr[i]-1]
temp1 = arr[i]
temp2 = arr[arr[i] - 1]
arr[i] = temp2
arr[temp1 - 1] = temp1
# Driver code
if __name__=='__main__':
arr = [ 2, 1, 5, 3, 4 ]
N = len(arr)
# Function call to sort the array
sortArray(arr, N)
# Function call to print the array
printArray(arr, N)
# This code is contributed by rutvik_56C#
// C# program for the above approach
using System;
class GFG{
// Function to print array element
public static void printArray(int []arr, int N)
{
// Traverse the array
for(int i = 0; i < N; i++)
{
Console.Write(arr[i] + " ");
}
}
// Function to sort the array in O(N)
public static void sortArray(int []arr, int N)
{
// Traverse the array
for(int i = 0; i < N; )
{
// If the current element is
// at correct position
if (arr[i] == i + 1)
{
i++;
}
// Else swap the current element
// with it's correct position
else
{
// Swap the value of
// arr[i] and arr[arr[i]-1]
int temp1 = arr[i];
int temp2 = arr[arr[i] - 1];
arr[i] = temp2;
arr[temp1 - 1] = temp1;
}
}
}
// Driver Code
public static void Main(String[] args)
{
int []arr = {2, 1, 5, 3, 4};
int N = arr.Length;
// Function call to sort the array
sortArray(arr, N);
// Function call to print the array
printArray(arr, N);
}
}
// This code is contributed by shikhasingrajputJavascript
输出：
1 2 3 4 5时间复杂度： O(N) ，其中 N 是数组的长度。
辅助空间： O(1)
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