import java.util.Arrays;

//利用二进制算法进行全排列

//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

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

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

result+=num[temp[i]];

}

System.out.println(result);

}

}

}

一.利用二进制状态法求排列组合，此种方法比较容易懂，但是运行效率不高，小数据排列组合可以使用

二.用递归的思想来求排列跟组合，代码量比较大

package practice;

import java.util.ArrayList;

import java.util.List;public classTest1 {/**

* @param args*/

public static voidmain(String[] args) {//TODO Auto-generated method stub

Object[] tmp={1,2,3,4,5,6};//ArrayList rs=RandomC(tmp);

ArrayList rs=cmn(tmp,3);for(int i=0;i

{//System.out.print(i+"=");

for(int j=0;j

{

System.out.print(rs.get(i)[j]+",");

}

System.out.println();

}

}//求一个数组的任意组合

static ArrayListRandomC(Object[] source)

{

ArrayList result=new ArrayList();if(source.length==1)

{

result.add(source);

}else{

Object[] psource=new Object[source.length-1];for(int i=0;i

{

psource[i]=source[i];

}

result=RandomC(psource);int len=result.size();//fn组合的长度

result.add((new Object[]{source[source.length-1]}));for(int i=0;i

{

Object[] tmp=new Object[result.get(i).length+1];for(int j=0;j

{

tmp[j]=result.get(i)[j];

}

tmp[tmp.length-1]=source[source.length-1];

result.add(tmp);

}

}returnresult;

}static ArrayList cmn(Object[] source,intn)

{

ArrayList result=new ArrayList();if(n==1)

{for(int i=0;i

{

result.add(newObject[]{source[i]});

}

}else if(source.length==n)

{

result.add(source);

}else{

Object[] psource=new Object[source.length-1];for(int i=0;i

{

psource[i]=source[i];

}

result=cmn(psource,n);

ArrayList tmp=cmn(psource,n-1);for(int i=0;i

{

Object[] rs=newObject[n];for(int j=0;j

{

rs[j]=tmp.get(i)[j];

}

rs[n-1]=source[source.length-1];

result.add(rs);

}

}returnresult;

}

}

三.利用动态规划的思想求排列和组合

packageAcm;//强大的求组合数

public classMainApp {public static voidmain(String[] args) {int[] num=new int[]{1,2,3,4,5};

String str="";//求3个数的组合个数//count(0,str,num,3);//求1-n个数的组合个数

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,intn) {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

packageAcm;//求排列，求各种排列或组合后排列

importjava.util.Arrays;importjava.util.Scanner;public classDemo19 {private static booleanf[];public static voidmain(String[] args) {

Scanner sc=newScanner(System.in);int sz=sc.nextInt();for(int i=0;i

f=new boolean[sum];

Arrays.fill(f,true);int[] num=new int[sum];for(int j=0;j

num[j]=j+1;

}int nn=sc.nextInt();

String str="";

count(num,str,nn);

}

}/***

*@paramnum 表示要排列的数组

*@paramstr 以排列好的字符串

*@paramnn 剩下需要排列的个数，如果需要全排列，则nn为数组长度*/

private static void count(int[] num, String str, intnn) {if(nn==0){

System.out.println(str);return;

}for(int i=0;i

}

f[i]=false;

count(num,str+num[i],nn-1);

f[i]=true;

}

}

}