java面试编程题、

给你一个地址，纯算法的，试试看：
http://topic.csdn.net/u/20100514/10/f7c34142-0718-4c59-8195-a711a8fd98e1.html如果算法的不算很牛看不懂的话，试试下面的全排列吧：public class PrintNumber {
public static void main(String[] args) {
List<String> list1 = new ArrayList<String>();
List<String> list2 = new ArrayList<String>();
for (int i = 1; i <= 5; i++) {
list1.add("" + i);
}
list1.add("" + 2);
while (list2.size() < 360) {
Collections.shuffle(list1);
StringBuffer sb = new StringBuffer();
for (int i = 0; i < list1.size(); i++) {
sb.append(list1.get(i));
}
if (list2.contains(sb.toString()) == false) {
list2.add(sb.toString());
}
}
int num = 0;
for (String str : list2) {
if (!str.matches("\\d\\d4\\d+") && !str.matches("\\d*35\\d*")
&& !str.matches("\\d*53\\d*")) {
System.out.println(str);
num++;
}
}
System.out.println("总共输入" + num + "个数"); }
}

import java.util.regex.Matcher;
import java.util.regex.Pattern;
import java.util.ArrayList;
import java.util.HashSet;public class Test{
public static void main(String[] args){
print();
} public static void print(){
String regex = "([1-6])(?!\\1)([1-6])(?!\\1|\\2|4)([1-6])(?!\\1|\\2|\\3)([1-6])(?!\\1|\\2|\\3|\\4)([1-6])(?!\\1|\\2|\\3|\\4|\\5)([1-6])$";
Pattern pattern = Pattern.compile(regex);
Matcher matcher = null;

HashSet<String> set = new HashSet<String>();
for(int i = 123456 ; i <= 654321 ; i ++){
String content = i + "";
matcher = pattern.matcher(content);
if(!matcher.matches()){
continue;
} String regex1 = "35";
content = content.replace("6","2");
Pattern pattern1 = Pattern.compile(regex1);
matcher = pattern1.matcher(content);
if(!matcher.find()){
set.add(content);
}
} System.out.println(set.size());
for(String s : set){
System.out.println(s);
}

}
}

用正则表达式想给个例子的，貌似java正则表达式还不会。。只在javascript中用了学习下再来

public class TestAllsort {
public static void main(String[] args) {
TestAllsort ta =new TestAllsort();
ta.sortAll(); }
public void sortAll(){
for(int k=0;k<100000;k++){
int [] arr = new int[5];
int [] b ={0,0,0,0,0};
for(int i=0;i<5;i++){
arr[i] = (int)((Math.random()*5)+1);
}
if(arr[2]==4){
continue;
}
for(int i=0;i<4;i++){
if(arr[i]==3&&arr[i+1]==5||arr[i]==5&&arr[i+1]==3){
continue;
}else{
for(int j=0;j<5;j++){
if(b[j]!=arr[j]){
b[j]=arr[j];
System.out.print(b[j]+" ");

}

}
}
System.out.println();
}
}
}
}
能达到要求，不过程序不简洁，呵呵只有这水平了。不要介意哈，还得加油

package com.wrox.context.chat;import java.util.HashSet;
import java.util.regex.Matcher;
import java.util.regex.Pattern;public class RexTest {

public static void eamilTest(String rouse){
String rex ="\\w+@\\w+[.]\\w+";
Pattern p = Pattern.compile(rex);
Matcher m = p.matcher(rouse);
if(m.find()){
System.out.println("符合要求");
}
else{
System.out.println("不符合要求");
}
} /**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
//Regex rex = new Regex();
//String url = "http://sina.com";
//String rex = "@http://.*?sina.*?\.com";
/* String parentString = "abcde";
// String sonString ="cd";
String rex ="[a-z].*cd.*";
Pattern p = Pattern.compile(rex);
Matcher m = p.matcher(parentString);
boolean b = m.find();
System.out.println(b);
*/
/*String root = "[email protected]";
//String rex = "\\w+@\\w+[.]\\w+";
eamilTest(root);*/
/* String s = "xxx==yyy==zzz==";
//String rex = "(?i)([^=]+)={2,}([^=]+)";
String rex = "(?i)([^=]+)={2,}([^=]+)";
// String rex = "(?i)//w+.*={2,}//w+"; Pattern p = Pattern.compile(rex);
Matcher m = p.matcher(s);
System.out.println(m.find());
*/

String rex ="[1-5][1-5][1-3[5]][1-5][1-5]";
Pattern p = Pattern.compile(rex);
Matcher m = null;

HashSet<String> hs = new HashSet<String>();
for(int i=12345;1<=54321;i++){
String context = i + "";
m=p.matcher(context);
if(!m.matches()){
continue;
}
String rex1="35|53";
Pattern p1 = Pattern.compile(rex1);
Matcher m1 = p1.matcher(context);
if(!m1.find()){
hs.add(context);
}
    for(String s:hs){
     System.out.println(s);
    }
}
}}

请问为什么这楼要判断if(b[j]!=arr[j])这个呢？是去除零还是什么？应该不会产生零吧？我不才指导下吧

static int[] bits = new int[] { 1, 2, 3, 4, 5 };
/**
* @param args
*/
public static void main(String[] args) {
sort("", bits);
}
private static void sort(String prefix, int[] a) {
if (a.length == 1) {
System.out.println(prefix + a[0]);
}
for (int i = 0; i < a.length; i++) {
sort(prefix + a, copy(a, i));
}
}
private static int[] copy(int[] a,int index){
int[] b = new int[a.length-1];
System.arraycopy(a, 0, b, 0, index);
System.arraycopy(a, index+1, b, index, a.length-index-1);
return b;
}

我觉得吧直接写个for循环，生成结果，再用那两条规则作下判断，所有结果存入hashMap前先判断有么有。
最后再输出。

6位全排列不算多。
System.out.println("512234");
System.out.println("412345");
...
跟考官说这是最快的方法。

不用怎么动脑的算法：public class test {
public static void main(String[] args) {
      long n = 0;
      for (int i0=1;i0<=6;i0++) //1
        for (int i1=1;i1<=6;i1++) //2
          for (int i2=1;i2<=6;i2++) //2
            if (i2>i1)
            for (int i3=1;i3<=6;i3++) //3
              for (int i4=1;i4<=6;i4++) //4
                if (i4!=3)
                for (int i5=1;i5<=6;i5++) //5
                  if ((i5!=i3+1)&&(i5!=i3-1)&&(i5-i4)*(i5-i3)*(i5-i2)*(i5-i1)*(i5-i0)*(i4-i3)*(i4-i2)*(i4-i1)*(i4-i0)*(i3-i2)*(i3-i1)*(i3-i0)*(i2-i0)*(i1-i0)!=0) {
                    n++;
                    char[] str = new char[6];
                    str[i0-1] = '1';
                    str[i1-1] = '2';
                    str[i2-1] = '2';
                    str[i3-1] = '3';
                    str[i4-1] = '4';
                    str[i5-1] = '5';
                    System.out.println("<"+n+"> ");
                    System.out.println(str);
                  } }
}

public class T12345 {

private int[] ns = {1, 2, 3, 4, 5}; private int nsIndex = 0;
/**
* 12345
* 21345
* 23145
* 23415
* 23451
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
new T12345().lunch1();

}

public void lunch() {
if(ns[ns.length-1] == 5) return;
if(ns[2]==4) {
moveNext();
lunch();
return;
}
if(l35()) {
return;
}
print();
moveNext();
lunch();
}
private boolean l35() {
// TODO Auto-generated method stub
int index = 0;
for (int i = 0; i < ns.length; i++) {
if(ns[i]==3||ns[i]==5) {
index = i;
break;
}
}
if(index == ns.length-1) return false;
if(index == 0) {
if(ns[index+1]==3||ns[index+1]==5) return true;
return false;
}
if(ns[index-1]==5||ns[index+1]==5||ns[index-1]==3||ns[index+1]==3) return true;
return false;
}
private void lunch1() {
// TODO Auto-generated method stub
if(ns[ns.length-1] == 1) {
lunch();
return;
}
if(ns[2]==4) {
moveNext();
lunch1();
return;
}
print();
moveNext();
lunch1();
} public void moveNext() {
if(nsIndex == ns.length-1) nsIndex = 0;
int temp = ns[nsIndex];
ns[nsIndex] = ns[nsIndex+1];
ns[nsIndex+1] = temp;
nsIndex++;

}

public void print() {
for(int n : ns) {
System.out.print(n);
}
System.out.println();
}}

String str = "[0-5][0-5][0-3[5]][0-5][0-5]";
             Pattern p = Pattern.compile(str);
             String context = "";
         for(int i=12345; i<54321;i++){
         context = i + "";
         Matcher m = p.matcher(context);
         if(!m.find()){
         continue;
         }
         String str2 = "35|53";
         Pattern p2 = Pattern.compile(str2);
         Matcher m2 = p2.matcher(context);
         if(!m2.find()){
         System.out.println(context);
         }

package test.play;import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;public class ListAll { /**
* 用1、2、2、3、4、5这六个数字，用java写一个main函数，打印出所有不同的排列，如：512234、412345等，要求："4"不能在第三位
* ，"3"与"5"不能相连。
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
char arr[] = { '1', '2', '2', '3', '4', '5' };
List<Integer> lst = new ArrayList<Integer>();
int min = 122345;
int max = 543221;
for (min = 122345; min <= max; min++) {
char temp[] = String.valueOf(min).toCharArray();
char mArr[] = Arrays.copyOf(temp, temp.length);
Arrays.sort(mArr);
if (new String(arr).equals(String.valueOf(mArr))
&& temp[2] != arr[4] && !String.valueOf(min).contains("35")
&& !String.valueOf(min).contains("53")) {
lst.add(min);
}
}
for (Integer i : lst) {
System.out.println(i);
}
}}

package com.chang.test;import java.util.ArrayList;
import java.util.List;public class TestCsdn1 { static List l = new ArrayList(); public TestCsdn1() { } public static void add(String n) {
List tmpl = new ArrayList(); for (int i = 0; i < l.size(); i++) {
String s = (String) l.get(i);
s = n + s;
tmpl.add(s);
String tmp = s;
s = change(s);
while (!s.equals(tmp)) {
tmpl.add(s);
s = change(s);
}
}
l.clear();
l.addAll(tmpl);
} private static String change(String s) {
char a[] = s.toCharArray();
char b = a[0];
s = s.substring(1, s.length()) + b;
return s;
} private static void printOut() {
int count = 0;
for (int i = 0; i < l.size(); i++) {
String s = (String) l.get(i);
if ((s.indexOf("3") + 1 == s.indexOf("5"))
|| (s.indexOf("3") - 1 == s.indexOf("5"))
|| (s.indexOf("4") == 2)) {
continue;
}
count++;
System.out.println(l.get(i));
}
System.out.println(l.size());
System.out.println(count); } public static void main(String[] args) { l.add("22");
add("1");
add("5");
add("4");
add("3"); printOut(); }}

我也贴一个我自己的
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;/**
* Title:全排列算法
*
* @author xxxxx
*/
public class FullArrayProblem {
//将NUM设置为待排列数组的长度即实现全排列
private int NUM = 6;
private int count;
private List<String> lresult;
/**
* 递归算法：将数据分为两部分，递归将数据从左侧移右侧实现全排列
*
* @param datas
* @param target
*/
private void sort(List<String> datas, StringBuffer target) {
if (target.length() == NUM) {
String strTarget = target.toString();
if (!lresult.contains(strTarget)&&!strTarget.matches("\\d\\d4\\d+")
&&!strTarget.matches("\\d*35\\d*")&&!strTarget.matches("\\d*53\\d*"))
{
System.out.println(target.toString());
lresult.add(strTarget);
count++;
}

return;
}
for (int i = 0; i < datas.size(); i++) {
List<String> newDatas = new ArrayList<String>(datas);
StringBuffer newTarget = new StringBuffer(target);
newTarget.append(newDatas.get(i));
newDatas.remove(i);
sort(newDatas, newTarget);
}
} public static void main(String[] args) {
String[] datas = new String[] { "1", "2", "2", "3" , "4" , "5" };
FullArrayProblem t = new FullArrayProblem();
t.lresult = new ArrayList<String>();
t.sort(Arrays.asList(datas), new StringBuffer());
System.out.println("TOTALNUMBER:" + t.count);
}}

private String[] b = new String[] { "1", "2", "2", "3", "4", "5" };
private int n = b.length;
private boolean[] visited = new boolean[n];
private int[][] a = new int[n][n];
private String result = "";
private TreeSet<String> set = new TreeSet<String>(); public static void main(String[] args) {
new RandomNum().start();
} private void start() {
// Initial the map a[][]
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (i == j) {
a[i][j] = 0;
} else {
a[i][j] = 1;
}
}
}
// 3 and 5 can not be the neighbor.
a[3][5] = 0;
a[5][3] = 0;
// Begin to depth　search.
for (int i = 0; i < n; i++) {
this.depthFirstSearch(i);
}
// Print result treeset.
Iterator<String> it = set.iterator();
while (it.hasNext()) {
String string = (String) it.next();
// "4" can not be the third position.
if (string.indexOf("4") != 2) {
System.out.println(string);
}
}
} private void depthFirstSearch(int startIndex) {
visited[startIndex] = true;
result = result + b[startIndex];
if (result.length() == n) {
// Filt the duplicate value.
set.add(result);
}
for (int j = 0; j < n; j++) {
if (a[startIndex][j] == 1 && visited[j] == false) {
depthFirstSearch(j);
} else {
continue;
}
}
// restore the result value and visited value after listing a node.
result = result.substring(0, result.length() - 1);
visited[startIndex] = false;
}

首先，自己封装了一个类，实现了全排列。这里设置了一个接口，要求继承它的子类，实现这个接口，以达到排列要求的限制
abstract class PaiLei { private int count;

public int count() {
return count;
}

/**
* 排列算法
* @param list 排序数据
* @param k 移动点
* @param m 数据个数
*/
public void paiLie(int list[], int k, int m) {
if (k == m) {
if (dataFilter(list)) {
count++;
System.out.println(Arrays.toString(list));
}
} else {
for (int i = k; i < m; i++) {
swap(list, k, i);
paiLie(list, k + 1, m);
swap(list, k, i);
}
}
}

/**
* 限制规则方法
* @param list 排列数组
* @return 是否通过
*/
public abstract boolean dataFilter(int[] list); private void swap(int[] datas, int i, int j) {
int t = datas[i];
datas[i] = datas[j];
datas[j] = t;
}
}根据题目的要求，写了一个测试类，并采用匿名类的方式，实现。
public static void main(String[] args) {
int list[] = { 1, 2, 2, 3, 4, 5};
PaiLei pl = new PaiLei() {

@Override
public boolean dataFilter(int[] list) {
int index = -1;
int n = list.length;
if (n > 3 && list[2] == 4) {
return false;
}
for (int i = 0; i < n; i++) {
if (list[i] == 3) {
index = i;
break;
}
}

return index == -1 ? true : !((index < n - 1 && list[index+1] == 5) ||
(index > 0 && list[index - 1] == 5));
}
};

pl.paiLie(list, 0, list.length);
System.out.println("count : " + pl.count());
}

public class Test {
static int count = 0;
static List list_all = new ArrayList();
static String regex = "35|53";
    static Pattern pattern = Pattern.compile(regex);
    /**
     * 递归
     */
public static List test(ArrayList list,String sbu){
for(int i=0;i<list.size();i++){
String sbu_temp = sbu + list.get(i);
     ArrayList list1 = (ArrayList) list.clone();
     list1.remove(i);
     if(list1.size()==0){
     Matcher matcher = pattern.matcher(sbu_temp);
     if(!list_all.contains(sbu_temp) && sbu_temp.indexOf("4")!=2 && !matcher.find()){
     System.out.println(sbu_temp);
     System.out.println(++count);
     list_all.add(sbu_temp);
     }
     }else{
     test(list1,sbu_temp);
     }

     }
return list_all;

}
    public static void main(String[] args) {
     ArrayList list = new ArrayList(6);
     list.add(1);
     list.add(2);
     list.add(2);
     list.add(3);
     list.add(4);
     list.add(5);
     test(list,"");
    }
}

题目如下：用1、2、2、3、4、5这六个数字，用java写一个main函数，打印出所有不同的排列，如：512234、412345等，要求："4"不能在第三位，"3"与"5"不能相连。

if (pre.length() == a.length) {
if (pre.charAt(2) != '4'&&!pre.contains("35")&&!pre.contains("53")) {
index++;
System.out.println(pre);
}
}
if (len < 1)
return;
for (int i = 0; i < len; i++) {
char temp = a[i];
swap(a, i, len - 1);
pailie(pre + temp, a, len - 1);
swap(a, len - 1, i);
}

调试易

java面试编程题、

解决方案 »