这道题怎么破？有好心人么？ - 黑马程序员技术交流社区

import java.io.*;
import java.util.*;
public class pllll {
public static void main(String[] args)throws Exception {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
String p=br.readLine();
yy(br,p );
}
public static void yy(BufferedReader br,String p ) throws Exception
{
int i,j; String ee=p; String r = null,t = null,q=null ;
List<String>list=new ArrayList<String>();
///////////////////////////////////////////////////////////////////
for(i=0;i<p.length();i++)
{
// sop(p.substring(i,i+1)+" ");////////////第一行
}
// System.out.println();
////////////////////////////////////////////////////////////////////
for(i=0;i<p.length();i++)
{
for(j=0;j<p.length();j++)
{
if(i!=j)
{
r=p.substring(i, i+1);
t=ee.substring(j,j+ 1);
q=r+t;
list.add(q);
//sop(q+" "); ////////////第二行
}
}
}
////////////////////////////////////////////////////////////////////
Iterator<String> iterator = list.iterator();
while(iterator.hasNext())
{
ee= iterator.next();
//sop(ee);
yy(ee,p);
}
}
public static void yy(String ee,String p) throws Exception
{ List<String>list=new ArrayList<String>();
int i,j; int k=2;
String r = null,t = null,q=null ;
for(i=0;i<p.length();i++)
{
r=p.substring(i, i+1);
t=ee.substring(0, k);//as ad sa sd da ds ,(0, 2)就是1,2位的，每个(as)(ad)....都是1，2位的即sop(t+" ");输出as ad sa sd da ds
q=r+t;
list.add(q);
sop(q+" ");
}
Iterator<String> iterator = list.iterator();
while(iterator.hasNext())
{
ee= iterator.next();
if(k<p.length()+1)
{ k++;
yy(ee,p) ;
}
}
}
public static void sop(Object obj)
{
System.out.print(obj);
}
}

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package com.mth.test;
/*
*上下对称
"a" "b" "c"
"ab" "bc" "ca"
"ba" "cb" "ac"
"abc" "bca" "cab"
"cba" "acb" "bac"
*
*
* 解题思路：把abc变成数组数组有下标根据下标组合来打印
*
* */
public class CharTest {
public static void main(String[] args) {
String string = "abc";
// 变成字符数组
char[] array = string.toCharArray();
//第一行
for (char a : array) {
System.out.print(a);
System.out.print(" ");
}
System.out.println();
//第二行
for (int i = 0; i < array.length; i++) {
if (i == array.length - 1) {
System.out.print(array[array.length - 1] + ""
+ array[array.length - 3]);
System.out.print(" ");
System.out.print(array[array.length - 3] + ""
+ array[array.length - 1]);
} else {
System.out.print(array[i] + "" + array[i + 1]);
System.out.print(" ");
System.out.print(array[i + 1] + "" + array[i]);
}
System.out.print(" ");
}
System.out.println();
//第三行
for (int i = 0; i < array.length; i++) {
if (i == 1) {
System.out.print(array[i] + "" + array[i + 1] + ""
+ array[i - 1]);
System.out.print(" ");
System.out.print(array[i - 1] + "" + array[i + 1] + ""
+ array[i]);
} else if (i == 2) {
System.out.print(array[i] + "" + array[i - 2] + ""
+ array[i - 1]);
System.out.print(" ");
System.out.print(array[i - 1] + "" + array[i - 2] + ""
+ array[i]);
} else {
System.out.print(array[i] + "" + array[i + 1] + ""
+ array[i + 2]);
System.out.print(" ");
System.out.print(array[i + 2] + "" + array[i + 1] + ""
+ array[i]);
}
System.out.print(" ");
}
}
}

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/*
* 需求：
* 编程列出一个字符串的全字符组合情况，原始字符串中没有重复字符
* 例：原始字符串是"abc"，打印得到下列所有组合情况
* "a" "b" "c"
* "ab" "bc" "ca" "ba" "cb" "ac"
* "abc" "acb" "bac" "bca" "cab" "cba"
*/
import java.io.*;
import java.util.*;
public class BasicTest
{
public static void main(String[] args) throws Exception
{
BufferedReader bufr = new BufferedReader(new InputStreamReader(System.in));
String line = bufr.readLine();
if(!valiStr(line))
{
System.out.println("输入的字符串中有重复的字符");
System.exit(0);
}
char[] charArr = line.toCharArray();
for(int x=1; x<=charArr.length; x++)
{
List<String> list = getCombineList(charArr, x);
System.out.println(x+"个字符的组合共有"+getCombineNumber(charArr, x)+"种组合：");
Iterator<String> it = list.iterator();
while(it.hasNext())
{
System.out.print(it.next()+" ");
}
System.out.println();
}
}
public static boolean valiStr(String str)
{
char[] arr = str.toCharArray();
List<String> list = new ArrayList<String>();
for(int i=0; i<arr.length; i++)
{
String tmp = new String(arr,i,1);
if(list.contains(tmp))
{
return false;
}
list.add(tmp);
}
return true;
}
public static int getCombineNumber(char[] charArr,int x)
{
int len = charArr.length;
int total = 1;
for(int i=0; i<x; i++)
{
total *= (len-i);
}
return total;
}
public static List<Integer> getIndexList(char[] charArr)
{
List<Integer> indexList = new ArrayList<Integer>();
for(int j=0; j<charArr.length; j++)
{
indexList.add(j);
}
return indexList;
}
public static String getRandomStr(char[] charArr,int x)
{
List<Integer> indexList = getIndexList(charArr);
int ilNum = indexList.size();
Random r = new Random();
char[] tmpArr = new char[x];
for(int i=0; i<x; i++)
{
int t = r.nextInt(ilNum-i);
tmpArr[i] = charArr[indexList.get(t)];
indexList.remove(t);
}
return new String(tmpArr);
}
public static List<String> getCombineList(char[] charArr,int x)
{
int total = getCombineNumber(charArr, x);
List<String> list = new ArrayList<String>();
while(true)
{
String str = getRandomStr(charArr, x);
if(list.contains(str))
continue;
list.add(str);
if(list.size() == total)
break;
}
return list;
}
}

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