打印所有可能的长度为 k 的字符串，这些字符串可以由一组 n 个字符组成
给定一组字符和一个正整数 k，打印所有可能的长度为 k 的字符串，这些字符串可以由给定的字符集组成。
示例:
Input:
set[] = {'a', 'b'}, k = 3
Output:
aaa
aab
aba
abb
baa
bab
bba
bbb
Input:
set[] = {'a', 'b', 'c', 'd'}, k = 1
Output:
a
b
c
d
对于给定的一组大小为 n 的字符串，将会有长度为 k 的 n^k 可能的字符串，其思想是从一个空的输出字符串开始(我们在下面的代码中称之为前缀)。逐个将所有字符添加到前缀中。对于添加的每个字符，通过递归调用等于 k-1 的 k 来打印当前前缀的所有可能的字符串。
以下是上述想法的实现:
C++
// C++ program to print all
// possible strings of length k
#include <bits/stdc++.h>
using namespace std;
// The main recursive method
// to print all possible
// strings of length k
void printAllKLengthRec(char set[], string prefix,
int n, int k)
{
// Base case: k is 0,
// print prefix
if (k == 0)
{
cout << (prefix) << endl;
return;
}
// One by one add all characters
// from set and recursively
// call for k equals to k-1
for (int i = 0; i < n; i++)
{
string newPrefix;
// Next character of input added
newPrefix = prefix + set[i];
// k is decreased, because
// we have added a new character
printAllKLengthRec(set, newPrefix, n, k - 1);
}
}
void printAllKLength(char set[], int k,int n)
{
printAllKLengthRec(set, "", n, k);
}
// Driver Code
int main()
{
cout << "First Test" << endl;
char set1[] = {'a', 'b'};
int k = 3;
printAllKLength(set1, k, 2);
cout << "Second Test\n";
char set2[] = {'a', 'b', 'c', 'd'};
k = 1;
printAllKLength(set2, k, 4);
}
// This code is contributed
// by Mohit kumar
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to print all
// possible strings of length k
class GFG {
// The method that prints all
// possible strings of length k.
// It is mainly a wrapper over
// recursive function printAllKLengthRec()
static void printAllKLength(char[] set, int k)
{
int n = set.length;
printAllKLengthRec(set, "", n, k);
}
// The main recursive method
// to print all possible
// strings of length k
static void printAllKLengthRec(char[] set,
String prefix,
int n, int k)
{
// Base case: k is 0,
// print prefix
if (k == 0)
{
System.out.println(prefix);
return;
}
// One by one add all characters
// from set and recursively
// call for k equals to k-1
for (int i = 0; i < n; ++i)
{
// Next character of input added
String newPrefix = prefix + set[i];
// k is decreased, because
// we have added a new character
printAllKLengthRec(set, newPrefix,
n, k - 1);
}
}
// Driver Code
public static void main(String[] args)
{
System.out.println("First Test");
char[] set1 = {'a', 'b'};
int k = 3;
printAllKLength(set1, k);
System.out.println("\nSecond Test");
char[] set2 = {'a', 'b', 'c', 'd'};
k = 1;
printAllKLength(set2, k);
}
}
Python 3
# Python 3 program to print all
# possible strings of length k
# The method that prints all
# possible strings of length k.
# It is mainly a wrapper over
# recursive function printAllKLengthRec()
def printAllKLength(set, k):
n = len(set)
printAllKLengthRec(set, "", n, k)
# The main recursive method
# to print all possible
# strings of length k
def printAllKLengthRec(set, prefix, n, k):
# Base case: k is 0,
# print prefix
if (k == 0) :
print(prefix)
return
# One by one add all characters
# from set and recursively
# call for k equals to k-1
for i in range(n):
# Next character of input added
newPrefix = prefix + set[i]
# k is decreased, because
# we have added a new character
printAllKLengthRec(set, newPrefix, n, k - 1)
# Driver Code
if __name__ == "__main__":
print("First Test")
set1 = ['a', 'b']
k = 3
printAllKLength(set1, k)
print("\nSecond Test")
set2 = ['a', 'b', 'c', 'd']
k = 1
printAllKLength(set2, k)
# This code is contributed
# by ChitraNayal
C
// C# program to print all
// possible strings of length k
using System;
class GFG {
// The method that prints all
// possible strings of length k.
// It is mainly a wrapper over
// recursive function printAllKLengthRec()
static void printAllKLength(char[] set, int k)
{
int n = set.Length;
printAllKLengthRec(set, "", n, k);
}
// The main recursive method
// to print all possible
// strings of length k
static void printAllKLengthRec(char[] set,
String prefix,
int n, int k)
{
// Base case: k is 0,
// print prefix
if (k == 0)
{
Console.WriteLine(prefix);
return;
}
// One by one add all characters
// from set and recursively
// call for k equals to k-1
for (int i = 0; i < n; ++i)
{
// Next character of input added
String newPrefix = prefix + set[i];
// k is decreased, because
// we have added a new character
printAllKLengthRec(set, newPrefix,
n, k - 1);
}
}
// Driver Code
static public void Main ()
{
Console.WriteLine("First Test");
char[] set1 = {'a', 'b'};
int k = 3;
printAllKLength(set1, k);
Console.WriteLine("\nSecond Test");
char[] set2 = {'a', 'b', 'c', 'd'};
k = 1;
printAllKLength(set2, k);
}
}
// This code is contributed by Ajit.
java 描述语言
<script>
// Javascript program to print all
// possible strings of length k
// The method that prints all
// possible strings of length k.
// It is mainly a wrapper over
// recursive function printAllKLengthRec()
function printAllKLength(set,k)
{
let n = set.length;
printAllKLengthRec(set, "", n, k);
}
// The main recursive method
// to print all possible
// strings of length k
function printAllKLengthRec(set,prefix,n,k)
{
// Base case: k is 0,
// print prefix
if (k == 0)
{
document.write(prefix+"<br>");
return;
}
// One by one add all characters
// from set and recursively
// call for k equals to k-1
for (let i = 0; i < n; ++i)
{
// Next character of input added
let newPrefix = prefix + set[i];
// k is decreased, because
// we have added a new character
printAllKLengthRec(set, newPrefix,
n, k - 1);
}
}
// Driver Code
document.write("First Test<br>");
let set1=['a', 'b'];
let k = 3;
printAllKLength(set1, k);
document.write("<br>Second Test<br>");
let set2 = ['a', 'b', 'c', 'd'];
k = 1;
printAllKLength(set2, k);
// This code is contributed by avanitrachhadiya2155
</script>
输出:
First Test
aaa
aab
aba
abb
baa
bab
bba
bbb
Second Test
a
b
c
d
以上解决方案主要是这个帖子的概括。
本文由 阿比纳夫·拉玛那 供稿。如果你发现任何不正确的地方，请写评论，或者你想分享更多关于上面讨论的话题的信息
版权属于：月萌API www.moonapi.com，转载请注明出处
