Java语言描述数学问题

/**

* @author eatonfang

* @version 1.0

*

*/

public class Test {

/**

* @param args

*/

public static void main(String[] args) {

// sample data for test

char[] a = new char[] {"a","b", "b", "c"};

int length = 4;

// process begin

// delete the same char

List<String> charList = new ArrayList<String>();

for (char temp : a) {

if (!charList.contains(String.valueOf(temp))) charList.add(String.valueOf(temp));

}

List<List<String>> arrayList = new ArrayList<List<String>>();

for (int i = 0; i < length-1; i++) {

arrayList.add(charList);

}

// begin zu he chu li

List<String> ressult = new ArrayList<String>();

List<String> subRessult = compose(arrayList);

for (String charValue : charList) {

// di gui method

for (String subStrResult : subRessult) {

ressult.add(charValue + subStrResult);

}

}

// show result

for (String temp : ressult) {

System.out.println(temp);

}

}

/**

* @param arrayList

*/

public static List<String> compose(List<List<String>> arrayList){

if (arrayList.size() == 1) return arrayList.get(0);

List<String> ressult = new ArrayList<String>();

List<String> temp = arrayList.get(0);

arrayList.remove(0);

List<String> subRessult = compose(arrayList);

for (String strTemp : temp) {

for (String subStrTemp : subRessult) {

ressult.add(strTemp + subStrTemp);

}

}

return ressult;

}

}

运行结果：

aaaa

aaab

aaac

aaba

aabb

aabc

aaca

aacb

aacc

abaa

abab

abac

abba

abbb

abbc

abca

abcb

abcc

acaa

acab

acac

acba

acbb

acbc

acca

accb

accc

baaa

baab

baac

baba

babb

babc

baca

bacb

bacc

bbaa

bbab

bbac

bbba

bbbb

bbbc

bbca

bbcb

bbcc

bcaa

bcab

bcac

bcba

bcbb

bcbc

bcca

bccb

bccc

caaa

caab

caac

caba

cabb

cabc

caca

cacb

cacc

cbaa

cbab

cbac

cbba

cbbb

cbbc

cbca

cbcb

cbcc

ccaa

ccab

ccac

ccba

ccbb

ccbc

ccca

cccb

cccc

我这里是的写法是防止你数组里面有重复的字符，导致出现重复的组合。然後你需要指定的是数组和想组合成多长的字符串，算法思想其实就是数学里面的排列与组合的思想，只不过用递归简化了写法

public class Du {

public static void main(String[] args) {

char[] ary = {"a", "b", "c"};

String str = new String(ary);

displayPermutation(str);

System.out.println("..............Duplicate starts.....");

repeatDuplicate(ary);

}

private static void repeatDuplicate(char[] ary) {

for(int i = 0; i < ary.length; i++){

StringBuilder sb = new StringBuilder();

for(int j = 0; j < ary.length; j++){

sb.append(ary[i]);

}

System.out.println(sb.toString());

}

}

public static void displayPermutation(String s) {

displayPermutation(" ", s);

}

public static void displayPermutation(String s1, String s2) {

if (s2.trim().equals("")) {

System.out.println(s1.trim());

} else {

StringBuilder sb = null;

for (int i = 0; i < s2.length(); i++) {

sb = new StringBuilder(s2);

String first = s1 + sb.charAt(i);//

displayPermutation(first, sb.deleteCharAt(i).toString());

}

}

}

}

----------------------------

abc

acb

bac

bca

cab

cba

..............Duplicate starts.....

aaa

bbb

ccc

import java.util.ArrayList;

import java.util.List;

public class ArrayPermutation {

public static void handler(char[] array){

List<char[]> valueList= new ArrayList<char[]>();

char[] value=new char[array.length];

int[] index=new int[array.length];

for(int i=0;i<array.length;i++){

value[i]=array[0];

}

while(index[array.length-1]<array.length){

valueList.add(value);

System.out.println(value);

index[0]++;

for(int i=0;i<array.length;i++){

if(index[i]==array.length){

if((i+1)!=array.length){

index[i]=0;

index[i+1]++;

value[i]=array[index[i]];

value[i+1]=array[index[i+1]%array.length];}

}

}

value[0]=array[index[0]];

}

}

public static void main(String[] args) {

handler(new char[]{"a","b","c"});

}

}

运行结果

aaa

baa

caa

aba

bba

cba

aca

bca

cca

aab

bab

cab

abb

bbb

cbb

acb

bcb

ccb

aac

bac

cac

abc

bbc

cbc

acc

bcc

ccc

楼上少了aab ,abb之类的两两重复的

先把需求搞清楚

字符可以重复出现，那对这个组合的长度有限定吗？如果没有那这个集合是无限的，