计算给定数字序列的可能解码

假设1代表“ A”，2代表“ B”，依此类推。给定一个数字序列，计算给定数字序列的可能解码次数。

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"

空数字序列被认为具有一个解码。可以假定输入包含从0到9的有效数字，并且没有前导0，没有多余的尾随0和没有两个或多个连续0。

此问题是递归的，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。我们重复两个子问题。
1)如果最后一位数字不为零，请重复计算剩余的(n-1)位数字并将结果加到总计数中。
2)如果最后两位数字构成有效字符(或小于27)，则重复剩余的(n-2)位数字并将结果加到总数中。

以下是上述方法的实现。

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 Sen

Python3

# 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 Hu

C#

// 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

 
// A naive recursive JavaScript implementation
// to count number of decodings that
// can be formed from a given digit sequence   
     
    // 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.
    function countDecoding(digits, 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
        let 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
    function countWays(digits, n)
    {
        if (n == 0 || (n == 1 && digits[0] == '0'))
        {
            return 0;
        }
        return countDecoding(digits, n);
    }
     
    // Driver code
    digits=['1', '2', '3', '4'];
    let n = digits.length;
    document.write("Count is ",countWays(digits, n));
     
    // This code is contributed by avanitrachhadiya2155
     

输出：

Count is 3

以上代码的时间复杂度是指数级的。如果我们仔细研究以上程序，可以发现递归解与斐波纳契数类似。因此，我们可以使用动态编程优化上述解决方案以使其在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

Python3

# 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

的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.
?>

Java脚本

 
// A Dynamic Programming based Javascript
// implementation to count decodings
 
// A Dynamic Programming based
// function to count decodings
function countDecodingDP(digits, n)
{
     
    // A table to store results of subproblems
    let count = new Array(n + 1);
    count[0] = 1;
    count[1] = 1;
     
    // For base condition "01123" should return 0
    if (digits[0] == '0')  
          return 0;
           
    for(let 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
let digits = [ '1','2','3','4' ];
let n = digits.length;
 
document.write("Count is " +
               countDecodingDP(digits, n));
                
// This code is contributed by rag2127
     

输出：

Count is 3