原文:
给定一个由个唯一元素组成的数组，我们必须使用数组的元素找到所有长度的排列 l 。允许重复元素。 例:
输入: arr = { 1，2 }，l=3 输出: 111 211 121 221 112 212 122 222 输入: arr = { 1，2，3 }，l=2 输出: 11
进场:
- 为了用 n 个元素形成长度为 l 的序列，已知序列的第i个元素可以用 n 种方式填充。所以会有序列
- 我们将运行一个从 0 到的循环，对于每个 i ，我们将把 i 从基地 10 转换为基地 n 。转换后的数字将代表数组的索引
- 我们可以通过这种方式打印所有的序列。
以下是实施办法:
c
// c implementation
#include
using namespace std;
// convert the number to lth
// base and print the sequence
void convert_to_len_th_base(int n,
int arr[],
int len,
int l)
{
// sequence is of length l
for (int i = 0; i < l; i ) {
// print the ith element
// of sequence
cout << arr[n % len];
n /= len;
}
cout << endl;
}
// print all the permuataions
void print(int arr[],
int len,
int l)
{
// there can be (len)^l
// permutations
for (int i = 0; i < (int)pow(len, l); i ) {
// convert i to len th base
convert_to_len_th_base(i, arr, len, l);
}
}
// driver code
int main()
{
int arr[] = { 1, 2, 3 };
int len = sizeof(arr) / sizeof(arr[0]);
int l = 2;
// function call
print(arr, len, l);
return 0;
}
java 语言(一种计算机语言，尤用于创建网站)
// java implementation for above approach
import java.io.*;
class gfg
{
// convert the number to lth
// base and print the sequence
static void convert_to_len_th_base(int n, int arr[],
int len, int l)
{
// sequence is of length l
for (int i = 0; i < l; i )
{
// print the ith element
// of sequence
system.out.print(arr[n % len]);
n /= len;
}
system.out.println();
}
// print all the permuataions
static void print(int arr[], int len, int l)
{
// there can be (len)^l
// permutations
for (int i = 0;
i < (int)math.pow(len, l); i )
{
// convert i to len th base
convert_to_len_th_base(i, arr, len, l);
}
}
// driver code
public static void main (string[] args)
{
int arr[] = { 1, 2, 3 };
int len = arr.length;
int l = 2;
// function call
print(arr, len, l);
}
}
// this code is contributed by ajit.
python 3
# python3 implementation for the above approach
# convert the number to lth
# base and print the sequence
def convert_to_len_th_base(n, arr, len, l):
# sequence is of length l
for i in range(l):
# print the ith element
# of sequence
print(arr[n % len], end = "")
n //= len
print()
# print all the permuataions
def printf(arr, len, l):
# there can be (len)^l permutations
for i in range(pow(len, l)):
# convert i to len th base
convert_to_len_th_base(i, arr, len, l)
# driver code
arr = [1, 2, 3]
len = len(arr)
l = 2
# function call
printf(arr, len, l)
# this code is contributed by mohit kumar
c
// c# implementation for above approach
using system;
class gfg
{
// convert the number to lth
// base and print the sequence
static void convert_to_len_th_base(int n, int []arr,
int len, int l)
{
// sequence is of length l
for (int i = 0; i < l; i )
{
// print the ith element
// of sequence
console.write(arr[n % len]);
n /= len;
}
console.writeline();
}
// print all the permuataions
static void print(int []arr, int len, int l)
{
// there can be (len)^l
// permutations
for (int i = 0;
i < (int)math.pow(len, l); i )
{
// convert i to len th base
convert_to_len_th_base(i, arr, len, l);
}
}
// driver code
public static void main (string[] args)
{
int []arr = { 1, 2, 3 };
int len = arr.length;
int l = 2;
// function call
print(arr, len, l);
}
}
// this code is contributed by rajput-ji
java 描述语言
output:
11
21
31
12
22
32
13
23
33
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