暴力破解密码破解Java
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Brute force password cracker Java
提问by Priyank
I am doing an assignment for class which I have to create a brute force password cracker in java.
我正在为班级做作业，我必须在 java 中创建一个蛮力密码破解程序。
- Write a function using Recursion to crack a password. The password is of unknown length (maximum 10) and is made up of capital letters and digits. (Store the actual password in your program, just for checking whether the string currently obtained is the right password.)
- 使用递归编写一个函数来破解密码。密码长度未知（最多 10 个），由大写字母和数字组成。（在你的程序中存储实际密码，只是为了检查当前获取的字符串是否是正确的密码。）
The solution I have is:
我的解决方案是：
import java.util.Arrays;
public class BruteForce {
public static void main(String[] args) {
bruteForce(3);
}
public static String bruteForce(int size) {
int[] password = new int[size];
String[] finalPassword = new String[size];
for (int i = 0; i < size; i++) {
password[i] = 0;
finalPassword[i] = "";
}
String pass = "AAA";
return computePermutations(size, password, 0, pass);
}
private static String computePermutations(int size, int[] password, int position, String pass) {
String testString = "";
String assemble = "";
for (int i = 0; i < 36; i++) {
password[position] = i;
if (position != size - 1) {
testString = computePermutations(size, password, position + 1, pass);
if (testString != "") {
return testString;
}
} else if (position == size - 1) {
for (int j = 0; j < size; j++) {
switch (password[j] + 1) {
case 1:
assemble = assemble + "A";
break;
case 2:
assemble = assemble + "B";
break;
case 3:
assemble = assemble + "C";
break;
case 4:
assemble = assemble + "D";
break;
case 5:
assemble = assemble + "E";
break;
case 6:
assemble = assemble + "F";
break;
case 7:
assemble = assemble + "G";
break;
case 8:
assemble = assemble + "H";
break;
case 9:
assemble = assemble + "I";
break;
case 10:
assemble = assemble + "J";
break;
case 11:
assemble = assemble + "K";
break;
case 12:
assemble = assemble + "L";
break;
case 13:
assemble = assemble + "M";
break;
case 14:
assemble = assemble + "N";
break;
case 15:
assemble = assemble + "O";
break;
case 16:
assemble = assemble + "P";
break;
case 17:
assemble = assemble + "Q";
break;
case 18:
assemble = assemble + "R";
break;
case 19:
assemble = assemble + "S";
break;
case 20:
assemble = assemble + "T";
break;
case 21:
assemble = assemble + "U";
break;
case 22:
assemble = assemble + "V";
break;
case 23:
assemble = assemble + "W";
break;
case 24:
assemble = assemble + "X";
break;
case 25:
assemble = assemble + "Y";
break;
case 26:
assemble = assemble + "Z";
break;
case 27:
assemble = assemble + "0";
break;
case 28:
assemble = assemble + "1";
break;
case 29:
assemble = assemble + "2";
break;
case 30:
assemble = assemble + "3";
break;
case 31:
assemble = assemble + "4";
break;
case 32:
assemble = assemble + "5";
break;
case 33:
assemble = assemble + "6";
break;
case 34:
assemble = assemble + "7";
break;
case 35:
assemble = assemble + "8";
break;
case 36:
assemble = assemble + "9";
break;
}
}
System.out.println(assemble);
if (assemble.equalsIgnoreCase(pass)) {
System.out.println("Password is: " + assemble);
break; //replace this with: return assemble;
} else {
assemble = "";
}
}
}
return "";
}
}
However, when I run the program, the password AAA never seems to work but 998 works just fine, what exactly is wrong?
但是，当我运行程序时，密码 AAA 似乎永远无法工作，而 998 却可以正常工作，究竟是什么问题？
采纳答案by Kartic
I think you should use a return statement if a match is found -
我认为如果找到匹配项，您应该使用 return 语句 -
if (assemble.equalsIgnoreCase(pass)) {
System.out.println("Password is: " + assemble);
return assemble; // This is missing
}
回答by Anirudh Jain
Your main problem seems to be to break the loop at the right point. Try using a label to break the loop and solve this issue. Find more about labels here: http://docs.oracle.com/javase/tutorial/java/nutsandbolts/branch.html
您的主要问题似乎是在正确的点打破循环。尝试使用标签打破循环并解决此问题。在此处查找有关标签的更多信息：http: //docs.oracle.com/javase/tutorial/java/nutsandbolts/branch.html
class BreakWithLabelDemo {
public static void main(String[] args) {
int[][] arrayOfInts = {
{ 32, 87, 3, 589 },
{ 12, 1076, 2000, 8 },
{ 622, 127, 77, 955 }
};
int searchfor = 12;
int i;
int j = 0;
boolean foundIt = false;
search:
for (i = 0; i < arrayOfInts.length; i++) {
for (j = 0; j < arrayOfInts[i].length;
j++) {
if (arrayOfInts[i][j] == searchfor) {
foundIt = true;
break search;
}
}
}
if (foundIt) {
System.out.println("Found " + searchfor + " at " + i + ", " + j);
} else {
System.out.println(searchfor + " not in the array");
}
}}
