给定一组字符,从中生成所有可能的密码。这意味着我们应该使用给定的字符,重新排列以及达到给定的长度来生成单词的所有可能排列。
例子:
Input : arr[] = {a, b},
len = 2.
Output :
a b aa ab ba bb
解决方案是在给定的字符数组上使用递归。这个想法是首先将所有可能的长度和一个空字符串传递给一个辅助函数。在辅助函数,我们将所有字符一一追加到当前字符串,然后重复填充剩余的字符串,直到达到所需的长度为止。
使用下面的递归树可以更好地可视化它:
(a, b)
/ \
a b
/ \ / \
aa ab ba bb
以下是上述方法的实现。
C++
// C++ program to generate all passwords for given characters
#include
using namespace std;
// int cnt;
// Recursive helper function, adds/removes characters
// until len is reached
void generate(char* arr, int i, string s, int len)
{
// base case
if (i == 0) // when len has been reached
{
// print it out
cout << s << "\n";
// cnt++;
return;
}
// iterate through the array
for (int j = 0; j < len; j++) {
// Create new string with next character
// Call generate again until string has
// reached its len
string appended = s + arr[j];
generate(arr, i - 1, appended, len);
}
return;
}
// function to generate all possible passwords
void crack(char* arr, int len)
{
// call for all required lengths
for (int i = 1; i <= len; i++) {
generate(arr, i, "", len);
}
}
// driver function
int main()
{
char arr[] = { 'a', 'b', 'c' };
int len = sizeof(arr) / sizeof(arr[0]);
crack(arr, len);
//cout << cnt << endl;
return 0;
}
// This code is contributed by Satish Srinivas.
Java
// Java program to generate all passwords for given characters
import java.util.*;
class GFG
{
// int cnt;
// Recursive helper function, adds/removes characters
// until len is reached
static void generate(char[] arr, int i, String s, int len)
{
// base case
if (i == 0) // when len has been reached
{
// print it out
System.out.println(s);
// cnt++;
return;
}
// iterate through the array
for (int j = 0; j < len; j++)
{
// Create new string with next character
// Call generate again until string has
// reached its len
String appended = s + arr[j];
generate(arr, i - 1, appended, len);
}
return;
}
// function to generate all possible passwords
static void crack(char[] arr, int len)
{
// call for all required lengths
for (int i = 1; i <= len; i++)
{
generate(arr, i, "", len);
}
}
// Driver code
public static void main(String[] args)
{
char arr[] = {'a', 'b', 'c'};
int len = arr.length;
crack(arr, len);
}
}
// This code has been contributed by 29AjayKumar
Python 3
# Python3 program to
# generate all passwords
# for given characters
# Recursive helper function,
# adds/removes characters
# until len is reached
def generate(arr, i, s, len):
# base case
if (i == 0): # when len has
# been reached
# print it out
print(s)
return
# iterate through the array
for j in range(0, len):
# Create new string with
# next character Call
# generate again until
# string has reached its len
appended = s + arr[j]
generate(arr, i - 1, appended, len)
return
# function to generate
# all possible passwords
def crack(arr, len):
# call for all required lengths
for i in range(1 , len + 1):
generate(arr, i, "", len)
# Driver Code
arr = ['a', 'b', 'c' ]
len = len(arr)
crack(arr, len)
# This code is contributed by Smita.
C#
// C# program to generate all passwords for given characters
using System;
class GFG
{
// int cnt;
// Recursive helper function, adds/removes characters
// until len is reached
static void generate(char[] arr, int i, String s, int len)
{
// base case
if (i == 0) // when len has been reached
{
// print it out
Console.WriteLine(s);
// cnt++;
return;
}
// iterate through the array
for (int j = 0; j < len; j++)
{
// Create new string with next character
// Call generate again until string has
// reached its len
String appended = s + arr[j];
generate(arr, i - 1, appended, len);
}
return;
}
// function to generate all possible passwords
static void crack(char[] arr, int len)
{
// call for all required lengths
for (int i = 1; i <= len; i++)
{
generate(arr, i, "", len);
}
}
// Driver code
public static void Main(String[] args)
{
char []arr = {'a', 'b', 'c'};
int len = arr.Length;
crack(arr, len);
}
}
/* This code contributed by PrinciRaj1992 */
Javascript
输出:
a
b
c
aa
ab
ac
ba
bb
bc
ca
cb
cc
aaa
aab
aac
aba
abb
abc
aca
acb
acc
baa
bab
bac
bba
bbb
bbc
bca
bcb
bcc
caa
cab
cac
cba
cbb
cbc
cca
ccb
ccc