The idea of full permutation and combination in java

1. Full arrangement

The idea of ​​full arrangement, in fact, full arrangement is a kind of arrangement and combination, such as 1 2 3, so how many possible combinations are there? Answer: A(3, 3)=3×2×1=6

• According to our subjective consciousness, there are naturally six kinds of ①1 2 3 ②1 3 2 ③2 1 3 ④2 3 1 ⑤3 1 2 ⑥3 2 1

• Next, explain the full arrangement idea in detail

• package test;
  
  import java.util.*;
  
  public class Main {
  	/*
  	 * 全排列的思想，其实全排列就是一种排列组合，比如 1 2 3，那么会有多少种组合的可能呢？
  	 *  以我们主观的意识，自然是有 ①1 2 3  ②1 3 2  ③2 1 3 ④2 3 1 ⑤3 1 2 ⑥3 2 1这六种
  	 *  接下来详解全排列思想
  	 * */
  	public static void main(String[] args) {
  		Scanner sc=new Scanner(System.in);
  		int n=sc.nextInt();  //输入n
  		int[] a=new int[n]; //创建数组a
  		for(int i=0;i<n;i++) {
  			a[i]=sc.nextInt(); // 获取数组a的元素
  		}
  		perm(a,0,n-1);
  	}
  	public static void perm(int[] a,int l,int r) { //l为左指针，r为右指针
  		if(l==r) System.out.println(Arrays.toString(a)); 
  		else {
  			for(int i=l;i<=r;i++) {
  				swap(a,i,l);  // 数组a[i]与a[l]作交换
  				perm(a,l+1,r); // 余下数组作全排列，固 左指针加1，即 l+1
  				swap(a,i,l); // 将数组还原防止重复
  			}
  		}
  	}
  	// 简单的作交换处理 ，定义 交换存储器 temp
  	public static void swap(int[] a,int i,int j) {
  		int temp = a[i];  
  		a[i]=a[j];
  		a[j]=temp;
  	}
  }
  

  

   

2. Combination

What is a combination, there is such a problem in mathematics, that is to take n balls out of m balls, how many combinations will there be, such as 1 2 3 4 four balls, take three out, the answer must be C ( 4, 3) = 4. The result is naturally ① 1 2 3 ② 1 2 4 ③ 2 3 4 ④ 1 3 4

Next, the combination idea will be explained in detail

package test;
import java.util.*;
import java.*;
import java.math.*;
public class Main { 
	static Stack<Integer> stack = new Stack<Integer>();
	static int cnt = 0;
	public static void main(String[] args) {
//	 
//	    从m个球里（编号为1,2,3...,m）一次取n个球，其中m>n，记录取出球的编号，枚举所有的可能性
	        int[] data = {1, 2, 3, 4};
	        recursion3(data, 0, 3, 0);
	        System.out.println(cnt);	
	} 
	public static void recursion3(int[] array, int curnum, int maxnum, int indexnum) {
        if (curnum == maxnum) {
            cnt++;
            System.out.println(stack);
            return;
        }
        for (int i = indexnum; i < array.length; i++) {
            if (!stack.contains(array[i])) {
                stack.push(array[i]);
                recursion3(array, curnum + 1, maxnum, i);
                stack.pop();
            }
        }
    }
	
}