java对数组进行排序_用Java对数组进行排序所需的最少交换

java对数组进行排序

Problem:

问题：

In this problem, we would have an unordered array with consecutive distinct natural numbers [1,2,3,..n], where n is the size of the array. We have to find the minimum number of swaps required to sort the array in ascending order.

在此问题中，我们将拥有一个具有连续的不同自然数[1,2,3，.. n]的无序数组，其中n是数组的大小。我们必须找到按升序对数组进行排序所需的最小交换次数 。

Note: Think and try by yourself before looking to the solution...

注意：在寻求解决方案之前，请自己思考并尝试...

Solution:

解：

This problem can be solved easily by observing the actual position of elements and their current position , the actual position of element in sorted array will be the a[cur]-1 (element-1), by tracking the actual position of element if we come back to the current element then there exist a cycle , then count the size of that cycle , the number of swaps will be cycling size-1, do this for all the cycles and add them together.

通过观察元素的实际位置及其当前位置，可以很容易地解决此问题，如果我们跟踪元素的实际位置，则排序数组中元素的实际位置将为a [cur -1 ] ( element-1 )回到当前元素，然后存在一个循环，然后计算该循环的大小，交换次数将为循环size-1 ，对所有循环进行此操作并将它们加在一起。

Example:

例：

Let an array: A =[2, 4, 5, 1, 3]

设一个数组：A = [2，4，5，1，3]

minimum swap required to sort an array in java

There exist two cycles:
Cycle 1: 2 → 4 → 1 → 2
Cycle 2: 5 → 3 → 5

存在两个周期：
周期1：2→4→1→2
周期2：5→3→5

Size of cycle = number of nodes:
Size of cycle 1=3
Size of cycle 2=2

周期大小=节点数：
周期1的大小= 3
周期2 = 2

Number of swaps: (3-1)+(2-1) = 3

掉期数量： (3-1)+(2-1)= 3

Program:

程序：

import java.io.*;
import java.math.*;
import java.util.*;
public class Swap {
static int minimumSwaps(int[] arr) {
int swap=0;
boolean visited[]=new boolean[arr.length];
for(int i=0;i<arr.length;i++){
int j=i,cycle=0;
while(!visited[j]){
visited[j]=true;
j=arr[j]-1;
cycle++;
}
if(cycle!=0)
swap+=cycle-1;
}
return swap;
}
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
int[] arr = new int[n];
for (int i = 0; i < n; i++) {
arr[i] = scanner.nextInt();
}
int res = minimumSwaps(arr);
System.out.println(res);
scanner.close();
}
}

Output

输出量