關於各種排列組合java演算法實現方法_java

一.利用二進位狀態法求排列組合，此種方法比較容易懂，但是運行效率不高，小資料排列組合可以使用

複製代碼 代碼如下:

import java.util.Arrays;

//利用二進位演算法進行全排列
//count1:170187
//count2:291656

public class test {
    public static void main(String[] args) {
        long start=System.currentTimeMillis();
        count2();
        long end=System.currentTimeMillis();
        System.out.println(end-start);
    }
    private static void count2(){
        int[] num=new int []{1,2,3,4,5,6,7,8,9};
        for(int i=1;i<Math.pow(9, 9);i++){
            String str=Integer.toString(i,9);
            int sz=str.length();
            for(int j=0;j<9-sz;j++){
                str="0"+str;
            }
            char[] temp=str.toCharArray();
            Arrays.sort(temp);
            String gl=new String(temp);
            if(!gl.equals("012345678")){
                continue;
            }
            String result="";
            for(int m=0;m<str.length();m++){
                result+=num[Integer.parseInt(str.charAt(m)+"")];
            }
            System.out.println(result);
        }
    }
    public static void count1(){
        int[] num=new int []{1,2,3,4,5,6,7,8,9};
        int[] ss=new int []{0,1,2,3,4,5,6,7,8};
        int[] temp=new int[9];
        while(temp[0]<9){
            temp[temp.length-1]++;
            for(int i=temp.length-1;i>0;i--){
                if(temp[i]==9){
                    temp[i]=0;
                    temp[i-1]++;
                }
            }
            int []tt=temp.clone();
            Arrays.sort(tt);
            if(!Arrays.equals(tt,ss)){
                continue;
            }
            String result="";
            for(int i=0;i<num.length;i++){
                result+=num[temp[i]];
            }
            System.out.println(result);

        }
    }
}

二.用遞迴的思想來求排列跟組合，代碼量比較大

複製代碼 代碼如下:

package practice;

import java.util.ArrayList;
import java.util.List;

public class Test1 {

    /**
     * @param args
     */
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        Object[] tmp={1,2,3,4,5,6};
//        ArrayList<Object[]> rs=RandomC(tmp);
        ArrayList<Object[]> rs=cmn(tmp,3);
        for(int i=0;i<rs.size();i++)
        {
//            System.out.print(i+"=");
            for(int j=0;j<rs.get(i).length;j++)
            {
                System.out.print(rs.get(i)[j]+",");
            }
            System.out.println();

        }
    }

   
    // 求一個數組的任意組合
    static ArrayList<Object[]> RandomC(Object[] source)
    {
        ArrayList<Object[]> result=new ArrayList<Object[]>();
        if(source.length==1)
        {
            result.add(source);       
        }
        else
        {
            Object[] psource=new Object[source.length-1];
            for(int i=0;i<psource.length;i++)
            {
                psource[i]=source[i];
            }
            result=RandomC(psource);
            int len=result.size();//fn組合的長度
            result.add((new Object[]{source[source.length-1]}));
            for(int i=0;i<len;i++)
            {
                Object[] tmp=new Object[result.get(i).length+1];
                for(int j=0;j<tmp.length-1;j++)
                {
                    tmp[j]=result.get(i)[j];
                }
                tmp[tmp.length-1]=source[source.length-1];
                result.add(tmp);
            }

        }
        return result;
    }

    static ArrayList<Object[]> cmn(Object[] source,int n)
    {
        ArrayList<Object[]> result=new ArrayList<Object[]>();
        if(n==1)
        {
            for(int i=0;i<source.length;i++)
            {
                result.add(new Object[]{source[i]});

            }
        }
        else if(source.length==n)
        {
            result.add(source);
        }
        else
        {
            Object[] psource=new Object[source.length-1];
            for(int i=0;i<psource.length;i++)
            {
                psource[i]=source[i];
            }
            result=cmn(psource,n);
            ArrayList<Object[]> tmp=cmn(psource,n-1);
            for(int i=0;i<tmp.size();i++)
            {
                Object[] rs=new Object[n];
                for(int j=0;j<n-1;j++)
                {
                    rs[j]=tmp.get(i)[j];
                }
                rs[n-1]=source[source.length-1];
                result.add(rs);
            }
        }
        return result;
    }

}

三.利用動態規劃的思想求排列和組合

複製代碼 代碼如下:

package Acm;
//強大的求組合數
public class MainApp {
    public static void main(String[] args) {
        int[] num=new int[]{1,2,3,4,5};
        String str="";
        //求3個數的組合個數
//        count(0,str,num,3);
//        求1-n個數的組合個數
        count1(0,str,num);
    }

    private static void count1(int i, String str, int[] num) {
        if(i==num.length){
            System.out.println(str);
            return;
        }
        count1(i+1,str,num);
        count1(i+1,str+num[i]+",",num);
    }

    private static void count(int i, String str, int[] num,int n) {
        if(n==0){
            System.out.println(str);
            return;
        }
        if(i==num.length){
            return;
        }
        count(i+1,str+num[i]+",",num,n-1);
        count(i+1,str,num,n);
    }
}

下面是求排列

複製代碼 代碼如下:

package Acm;
//求排列，求各種排列或組合後排列
import java.util.Arrays;
import java.util.Scanner;

public class Demo19 {
    private static boolean f[];
    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        int sz=sc.nextInt();
        for(int i=0;i<sz;i++){
            int sum=sc.nextInt();
            f=new boolean[sum];
            Arrays.fill(f, true);
            int[] num=new int[sum];
            for(int j=0;j<sum;j++){
                num[j]=j+1;
            }
            int nn=sc.nextInt();
            String str="";
            count(num,str,nn);
        }
    }
    /**
     *
     * @param num 表示要排列的數組
     * @param str 以排列好的字串
     * @param nn  剩下需要排列的個數，如果需要全排列，則nn為數組長度
     */
    private static void count(int[] num, String str, int nn) {
        if(nn==0){
            System.out.println(str);
            return;
        }
        for(int i=0;i<num.length;i++){
            if(!f[i]){
                continue;
            }
            f[i]=false;
            count(num,str+num[i],nn-1);
            f[i]=true;
        }
    }
}