让1代表’A’，2代表’B’等。给定一个数字序列，计算给定数字序列可能的解码次数。
例子：
Input: digits[] = "121"
Output: 3
// The possible decodings are "ABA", "AU", "LA"
Input: digits[] = "1234"
Output: 3
// The possible decodings are "ABCD", "LCD", "AWD"一个空的数字序列被认为有一个解码。可以假设输入包含从 0 到 9 的有效数字，并且没有前导 0，没有额外的尾随 0，并且没有两个或多个连续的 0。
这个问题是递归的，可以分解为子问题。我们从给定数字序列的末尾开始。我们将解码的总数初始化为 0。我们对两个子问题重复出现。
1) 如果最后一位数字不为零，则对剩余的 (n-1) 位数字重复并将结果添加到总数中。
2) 如果最后两位数字构成一个有效字符(或小于 27)，则重复剩余的 (n-2) 位数字并将结果添加到总数中。
下面是上述方法的实现。
C++
// C++ implementation to count number of
// decodings that can be formed from a
// given digit sequence
#include
#include
using namespace std;
// recuring function to find
// ways in how many ways a
// string can be decoded of length
// greater than 0 and starting with
// digit 1 and greater.
int countDecoding(char* digits, int n)
{
// base cases
if (n == 0 || n == 1)
return 1;
if (digits[0] == '0')
return 0;
// for base condition "01123" should return 0
// Initialize count
int count = 0;
// If the last digit is not 0,
// then last digit must add
// to the number of words
if (digits[n - 1] > '0')
count = countDecoding(digits, n - 1);
// If the last two digits form a number smaller
// than or equal to 26, then consider
// last two digits and recur
if (digits[n - 2] == '1'
|| (digits[n - 2] == '2'
&& digits[n - 1] < '7'))
count += countDecoding(digits, n - 2);
return count;
}
// Given a digit sequence of length n,
// returns count of possible decodings by
// replacing 1 with A, 2 woth B, ... 26 with Z
int countWays(char* digits, int n)
{
if (n == 0 || (n == 1 && digits[0] == '0'))
return 0;
return countDecoding(digits, n);
}
// Driver code
int main()
{
char digits[] = "1234";
int n = strlen(digits);
cout << "Count is " << countWays(digits, n);
return 0;
}
// Modified by Atanu Sen Java
// A naive recursive Java implementation
// to count number of decodings that
// can be formed from a given digit sequence
class GFG {
// recuring function to find
// ways in how many ways a
// string can be decoded of length
// greater than 0 and starting with
// digit 1 and greater.
static int countDecoding(char[] digits, int n)
{
// base cases
if (n == 0 || n == 1)
return 1;
// for base condition "01123" should return 0
if (digits[0] == '0')
return 0;
// Initialize count
int count = 0;
// If the last digit is not 0, then
// last digit must add to
// the number of words
if (digits[n - 1] > '0')
count = countDecoding(digits, n - 1);
// If the last two digits form a number
// smaller than or equal to 26,
// then consider last two digits and recur
if (digits[n - 2] == '1'
|| (digits[n - 2] == '2'
&& digits[n - 1] < '7'))
count += countDecoding(digits, n - 2);
return count;
}
// Given a digit sequence of length n,
// returns count of possible decodings by
// replacing 1 with A, 2 woth B, ... 26 with Z
static int countWays(char[] digits, int n)
{
if (n == 0 || (n == 1 && digits[0] == '0'))
return 0;
return countDecoding(digits, n);
}
// Driver code
public static void main(String[] args)
{
char digits[] = { '1', '2', '3', '4' };
int n = digits.length;
System.out.printf("Count is %d",
countWays(digits, n));
}
}
// This code is contributed by Smitha Dinesh Semwal.
// Modified by Atanu SenPython3
# Recursive implementation of numDecodings
def numDecodings(s: str) -> int:
if len(s) == 0
or (len(s) == 1
and s[0] == '0'):
return 0
return numDecodingsHelper(s, len(s))
def numDecodingsHelper(s: str, n: int) -> int:
if n == 0 or n == 1:
return 1
count = 0
if s[n-1] > "0":
count = numDecodingsHelper(s, n-1)
if (s[n - 2] == '1'
or (s[n - 2] == '2'
and s[n - 1] < '7')):
count += numDecodingsHelper(s, n - 2)
return count
# Driver code
digits = "1234"
print("Count is ", numDecodings(digits))
# This code is contributed by Frank HuC#
// A naive recursive C# implementation
// to count number of decodings that
// can be formed from a given digit sequence
using System;
class GFG {
// recuring function to find
// ways in how many ways a
// string can be decoded of length
// greater than 0 and starting with
// digit 1 and greater.
static int countDecoding(char[] digits, int n)
{
// base cases
if (n == 0 || n == 1)
return 1;
// Initialize count
int count = 0;
// If the last digit is not 0, then
// last digit must add to
// the number of words
if (digits[n - 1] > '0')
count = countDecoding(digits, n - 1);
// If the last two digits form a number
// smaller than or equal to 26, then
// consider last two digits and recur
if (digits[n - 2] == '1'
|| (digits[n - 2] == '2'
&& digits[n - 1] < '7'))
count += countDecoding(digits, n - 2);
return count;
}
// Given a digit sequence of length n,
// returns count of possible decodings by
// replacing 1 with A, 2 woth B, ... 26 with Z
static int countWays(char[] digits, int n)
{
if (n == 0 || (n == 1 && digits[0] == '0'))
return 0;
return countDecoding(digits, n);
}
// Driver code
public static void Main()
{
char[] digits = { '1', '2', '3', '4' };
int n = digits.Length;
Console.Write("Count is ");
Console.Write(countWays(digits, n));
}
}
// This code is contributed by nitin mittal.PHP
'0')
$count = countDecoding($digits, $n - 1);
// If the last two digits form a number
// smaller than or equal to 26, then
// consider last two digits and recur
if ($digits[$n - 2] == '1' ||
($digits[$n - 2] == '2' &&
$digits[$n - 1] < '7') )
$count += countDecoding($digits, $n - 2);
return $count;
}
// Given a digit sequence of length n,
// returns count of possible decodings by
// replacing 1 with A, 2 woth B, ... 26 with Z
function countWays(&$digits, $n){
if($n==0 || ($n == 1 && $digits[0] == '0'))
return 0;
return countDecoding($digits, $n);
}
// Driver Code
$digits = "1234";
$n = strlen($digits);
echo "Count is " . countWays($digits, $n);
// This code is contributed by ita_c
?>Javascript
C++
// A Dynamic Programming based C++
// implementation to count decodings
#include
#include
using namespace std;
// A Dynamic Programming based function
// to count decodings
int countDecodingDP(char *digits, int n)
{
// A table to store results of subproblems
int count[n+1];
count[0] = 1;
count[1] = 1;
//for base condition "01123" should return 0
if(digits[0]=='0')
return 0;
for (int i = 2; i <= n; i++)
{
count[i] = 0;
// If the last digit is not 0,
// then last digit must add to the number of words
if (digits[i-1] > '0')
count[i] = count[i-1];
// If second last digit is smaller
// than 2 and last digit is smaller than 7,
// then last two digits form a valid character
if (digits[i-2] == '1' ||
(digits[i-2] == '2' && digits[i-1] < '7') )
count[i] += count[i-2];
}
return count[n];
}
// Driver program to test above function
int main()
{
char digits[] = "1234";
int n = strlen(digits);
cout << "Count is " << countDecodingDP(digits, n);
return 0;
}
// Modified by Atanu Sen Java
// A Dynamic Programming based Java
// implementation to count decodings
import java.io.*;
class GFG
{
// A Dynamic Programming based
// function to count decodings
static int countDecodingDP(char digits[],
int n)
{
// A table to store results of subproblems
int count[] = new int[n + 1];
count[0] = 1;
count[1] = 1;
if(digits[0]=='0') //for base condition "01123" should return 0
return 0;
for (int i = 2; i <= n; i++)
{
count[i] = 0;
// If the last digit is not 0,
// then last digit must add to
// the number of words
if (digits[i - 1] > '0')
count[i] = count[i - 1];
// If second last digit is smaller
// than 2 and last digit is smaller
// than 7, then last two digits
// form a valid character
if (digits[i - 2] == '1' ||
(digits[i - 2] == '2' &&
digits[i - 1] < '7'))
count[i] += count[i - 2];
}
return count[n];
}
// Driver Code
public static void main (String[] args)
{
char digits[] = {'1','2','3','4'};
int n = digits.length;
System.out.println("Count is " +
countDecodingDP(digits, n));
}
}
// This code is contributed by anuj_67
// Modified by Atanu SenPython3
# A Dynamic Programming based Python3
# implementation to count decodings
# A Dynamic Programming based function
# to count decodings
def countDecodingDP(digits, n):
count = [0] * (n + 1); # A table to store
# results of subproblems
count[0] = 1;
count[1] = 1;
for i in range(2, n + 1):
count[i] = 0;
# If the last digit is not 0, then last
# digit must add to the number of words
if (digits[i - 1] > '0'):
count[i] = count[i - 1];
# If second last digit is smaller than 2
# and last digit is smaller than 7, then
# last two digits form a valid character
if (digits[i - 2] == '1' or
(digits[i - 2] == '2' and
digits[i - 1] < '7') ):
count[i] += count[i - 2];
return count[n];
# Driver Code
digits = "1234";
n = len(digits);
print("Count is" ,
countDecodingDP(digits, n));
# This code is contributed by mitsC#
// A Dynamic Programming based C#
// implementation to count decodings
using System;
class GFG
{
// A Dynamic Programming based
// function to count decodings
static int countDecodingDP(char[] digits,
int n)
{
// A table to store results of subproblems
int[] count = new int[n + 1];
count[0] = 1;
count[1] = 1;
for (int i = 2; i <= n; i++)
{
count[i] = 0;
// If the last digit is not 0,
// then last digit must add to
// the number of words
if (digits[i - 1] > '0')
count[i] = count[i - 1];
// If second last digit is smaller
// than 2 and last digit is smaller
// than 7, then last two digits
// form a valid character
if (digits[i - 2] == '1' ||
(digits[i - 2] == '2' &&
digits[i - 1] < '7'))
count[i] += count[i - 2];
}
return count[n];
}
// Driver Code
public static void Main()
{
char[] digits = {'1','2','3','4'};
int n = digits.Length;
Console.WriteLine("Count is " +
countDecodingDP(digits, n));
}
}
// This code is contributed
// by Akanksha Rai
// Modified by Atanu SenPHP
'0')
$count[$i] = $count[$i-1];
// If second last digit is smaller than 2 and last digit is
// smaller than 7, then last two digits form a valid character
if ($digits[$i-2] == '1' || ($digits[$i-2] == '2' && $digits[$i-1] < '7') )
$count[$i] += $count[$i-2];
}
return $count[$n];
}
// Driver program to test above function
$digits = "1234";
$n = strlen($digits);
echo "Count is " , countDecodingDP($digits, $n);
#This code is contributed by ajit.
?>Javascript
输出：
Count is 3上述代码的时间复杂度是指数级的。如果我们仔细查看上面的程序，我们可以观察到递归解决方案类似于斐波那契数列。因此，我们可以使用动态规划优化上述解决方案以在 O(n) 时间内工作。
以下是相同的实现。
C++
// A Dynamic Programming based C++
// implementation to count decodings
#include
#include
using namespace std;
// A Dynamic Programming based function
// to count decodings
int countDecodingDP(char *digits, int n)
{
// A table to store results of subproblems
int count[n+1];
count[0] = 1;
count[1] = 1;
//for base condition "01123" should return 0
if(digits[0]=='0')
return 0;
for (int i = 2; i <= n; i++)
{
count[i] = 0;
// If the last digit is not 0,
// then last digit must add to the number of words
if (digits[i-1] > '0')
count[i] = count[i-1];
// If second last digit is smaller
// than 2 and last digit is smaller than 7,
// then last two digits form a valid character
if (digits[i-2] == '1' ||
(digits[i-2] == '2' && digits[i-1] < '7') )
count[i] += count[i-2];
}
return count[n];
}
// Driver program to test above function
int main()
{
char digits[] = "1234";
int n = strlen(digits);
cout << "Count is " << countDecodingDP(digits, n);
return 0;
}
// Modified by Atanu Sen
Java
// A Dynamic Programming based Java
// implementation to count decodings
import java.io.*;
class GFG
{
// A Dynamic Programming based
// function to count decodings
static int countDecodingDP(char digits[],
int n)
{
// A table to store results of subproblems
int count[] = new int[n + 1];
count[0] = 1;
count[1] = 1;
if(digits[0]=='0') //for base condition "01123" should return 0
return 0;
for (int i = 2; i <= n; i++)
{
count[i] = 0;
// If the last digit is not 0,
// then last digit must add to
// the number of words
if (digits[i - 1] > '0')
count[i] = count[i - 1];
// If second last digit is smaller
// than 2 and last digit is smaller
// than 7, then last two digits
// form a valid character
if (digits[i - 2] == '1' ||
(digits[i - 2] == '2' &&
digits[i - 1] < '7'))
count[i] += count[i - 2];
}
return count[n];
}
// Driver Code
public static void main (String[] args)
{
char digits[] = {'1','2','3','4'};
int n = digits.length;
System.out.println("Count is " +
countDecodingDP(digits, n));
}
}
// This code is contributed by anuj_67
// Modified by Atanu Sen
蟒蛇3
# A Dynamic Programming based Python3
# implementation to count decodings
# A Dynamic Programming based function
# to count decodings
def countDecodingDP(digits, n):
count = [0] * (n + 1); # A table to store
# results of subproblems
count[0] = 1;
count[1] = 1;
for i in range(2, n + 1):
count[i] = 0;
# If the last digit is not 0, then last
# digit must add to the number of words
if (digits[i - 1] > '0'):
count[i] = count[i - 1];
# If second last digit is smaller than 2
# and last digit is smaller than 7, then
# last two digits form a valid character
if (digits[i - 2] == '1' or
(digits[i - 2] == '2' and
digits[i - 1] < '7') ):
count[i] += count[i - 2];
return count[n];
# Driver Code
digits = "1234";
n = len(digits);
print("Count is" ,
countDecodingDP(digits, n));
# This code is contributed by mits
C＃
// A Dynamic Programming based C#
// implementation to count decodings
using System;
class GFG
{
// A Dynamic Programming based
// function to count decodings
static int countDecodingDP(char[] digits,
int n)
{
// A table to store results of subproblems
int[] count = new int[n + 1];
count[0] = 1;
count[1] = 1;
for (int i = 2; i <= n; i++)
{
count[i] = 0;
// If the last digit is not 0,
// then last digit must add to
// the number of words
if (digits[i - 1] > '0')
count[i] = count[i - 1];
// If second last digit is smaller
// than 2 and last digit is smaller
// than 7, then last two digits
// form a valid character
if (digits[i - 2] == '1' ||
(digits[i - 2] == '2' &&
digits[i - 1] < '7'))
count[i] += count[i - 2];
}
return count[n];
}
// Driver Code
public static void Main()
{
char[] digits = {'1','2','3','4'};
int n = digits.Length;
Console.WriteLine("Count is " +
countDecodingDP(digits, n));
}
}
// This code is contributed
// by Akanksha Rai
// Modified by Atanu Sen
PHP
'0')
$count[$i] = $count[$i-1];
// If second last digit is smaller than 2 and last digit is
// smaller than 7, then last two digits form a valid character
if ($digits[$i-2] == '1' || ($digits[$i-2] == '2' && $digits[$i-1] < '7') )
$count[$i] += $count[$i-2];
}
return $count[$n];
}
// Driver program to test above function
$digits = "1234";
$n = strlen($digits);
echo "Count is " , countDecodingDP($digits, $n);
#This code is contributed by ajit.
?>
Javascript
输出：
Count is 3如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。