给定一个由 1 到 N 的不同整数组成的数组arr[] 。任务是找到对数组进行排序所需的最小交换次数。
例子：
Input: arr[] = { 7, 1, 3, 2, 4, 5, 6 }
Output: 5
Explanation:
i arr swap (indices)
0 [7, 1, 3, 2, 4, 5, 6] swap (0, 3)
1 [2, 1, 3, 7, 4, 5, 6] swap (0, 1)
2 [1, 2, 3, 7, 4, 5, 6] swap (3, 4)
3 [1, 2, 3, 4, 7, 5, 6] swap (4, 5)
4 [1, 2, 3, 4, 5, 7, 6] swap (5, 6)
5 [1, 2, 3, 4, 5, 6, 7]
Therefore, total number of swaps = 5
Input: arr[] = { 2, 3, 4, 1, 5 }
Output: 3方法：
- 对于arr[] 中的每个索引。
- 检查当前元素是否在正确的位置。由于数组包含从 1 到 N 的不同元素，我们可以简单地将元素与其在数组中的索引进行比较，以检查它是否在正确的位置。
- 如果当前元素不在其正确位置，则将元素与已占据其位置的元素交换。
- 否则检查下一个索引。
下面是上述方法的实现：
C++
#include
using namespace std;
// Function to find minimum swaps
int minimumSwaps(int arr[],int n)
{
// Initialise count variable
int count = 0;
int i = 0;
while (i < n)
{
// If current element is
// not at the right position
if (arr[i] != i + 1)
{
while (arr[i] != i + 1)
{
int temp = 0;
// Swap current element
// with correct position
// of that element
temp = arr[arr[i] - 1];
arr[arr[i] - 1] = arr[i];
arr[i] = temp;
count++;
}
}
// Increment for next index
// when current element is at
// correct position
i++;
}
return count;
}
// Driver code
int main()
{
int arr[] = { 2, 3, 4, 1, 5 };
int n = sizeof(arr)/sizeof(arr[0]);
// Function to find minimum swaps
cout << minimumSwaps(arr,n) ;
}
// This code is contributed by AnkitRai01 Java
// Java program to find the minimum
// number of swaps required to sort
// the given array
import java.io.*;
import java.util.*;
class GfG {
// Function to find minimum swaps
static int minimumSwaps(int[] arr)
{
// Initialise count variable
int count = 0;
int i = 0;
while (i < arr.length) {
// If current element is
// not at the right position
if (arr[i] != i + 1) {
while (arr[i] != i + 1) {
int temp = 0;
// Swap current element
// with correct position
// of that element
temp = arr[arr[i] - 1];
arr[arr[i] - 1] = arr[i];
arr[i] = temp;
count++;
}
}
// Increment for next index
// when current element is at
// correct position
i++;
}
return count;
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 2, 3, 4, 1, 5 };
// Function to find minimum swaps
System.out.println(minimumSwaps(arr));
}
}Python3
# Python3 program to find the minimum
# number of swaps required to sort
# the given array
# Function to find minimum swaps
def minimumSwaps(arr):
# Initialise count variable
count = 0;
i = 0;
while (i < len(arr)):
# If current element is
# not at the right position
if (arr[i] != i + 1):
while (arr[i] != i + 1):
temp = 0;
# Swap current element
# with correct position
# of that element
temp = arr[arr[i] - 1];
arr[arr[i] - 1] = arr[i];
arr[i] = temp;
count += 1;
# Increment for next index
# when current element is at
# correct position
i += 1;
return count;
# Driver code
if __name__ == '__main__':
arr = [ 2, 3, 4, 1, 5 ];
# Function to find minimum swaps
print(minimumSwaps(arr));
# This code is contributed by 29AjayKumarC#
// C# program to find the minimum
// number of swaps required to sort
// the given array
using System;
class GfG
{
// Function to find minimum swaps
static int minimumSwaps(int[] arr)
{
// Initialise count variable
int count = 0;
int i = 0;
while (i < arr.Length)
{
// If current element is
// not at the right position
if (arr[i] != i + 1)
{
while (arr[i] != i + 1)
{
int temp = 0;
// Swap current element
// with correct position
// of that element
temp = arr[arr[i] - 1];
arr[arr[i] - 1] = arr[i];
arr[i] = temp;
count++;
}
}
// Increment for next index
// when current element is at
// correct position
i++;
}
return count;
}
// Driver code
public static void Main(String[] args)
{
int []arr = { 2, 3, 4, 1, 5 };
// Function to find minimum swaps
Console.WriteLine(minimumSwaps(arr));
}
}
// This code is contributed by 29AjayKumarJavascript
输出：
3时间复杂度：O(N)，其中 N 是数组的大小。
辅助空间： O(1)
如果您想与行业专家一起参加直播课程，请参阅Geeks Classes Live