在 O(N) 时间和恒定辅助空间中计算给定数字序列的可能解码

给定一个数字序列S ，任务是找到给定数字序列的可能解码次数，其中 1 代表“A”，2 代表“B”……等等，直到 26，其中 26 代表“Z”。

例子：

Input: S = “121”
Output: 3
The possible decodings are “ABA”, “AU”, “LA”

Input: S = “1234”
Output: 3
The possible decodings are “ABCD”, “LCD”, “AWD”

方法：为了以O(N)的时间复杂度解决这个问题，使用动态规划。并且为了将辅助空间复杂度降低到 O(1)，我们使用了斐波那契数帖中讨论的递推关系的空间优化版本。
与斐波那契数列类似，任何当前“^第i^个”指数的关键观察值都可以使用其前两个指数进行计算。所以计算第 i^个索引的递归关系可以表示为

// Condition to check last
// digit can be included or not
if (digit[i-1] is not '0')
     count[i] += count[i-1]

// Condition to check the last
// two digits contribution
if (digit[i-2] is 1 or 
   (digit[i-2] is 2 and 
    digit[i-1] is less than 7))
     count[i] += count[i-2]

下面是上述方法的实现：

C++

// C++ implementation to count decodings
 
#include 
using namespace std;
 
// A Dynamic Programming based function
// to count decodings in digit sequence
int countDecodingDP(string digits, int n)
{
    // For base condition "01123"
    // should return 0
    if (digits[0] == '0')
        return 0;
 
    int count0 = 1, count1 = 1, count2;
 
    // Using last two calculated values,
    // calculate for ith index
    for (int i = 2; i <= n; i++) {
        count2 = ((int)(digits[i - 1] != '0') *  count1) +
                  (int)((digits[i - 2] == '1') or
                  (digits[i - 2] == '2' and
                   digits[i - 1] < '7')) * count0;
        count0 = count1;
        count1 = count2;
    }
 
    // Return the required answer
    return count1;
}
 
// Driver Code
int main()
{
    string digits = "1234";
    int n = digits.size();
 
    // Function call
    cout << countDecodingDP(digits, n);
 
    return 0;
}

Java

// Java implementation to count decodings
class GFG{
 
// A Dynamic programming based function
// to count decodings in digit sequence
static int countDecodingDP(String digits, int n)
{
     
    // For base condition "01123"
    // should return 0
    if (digits.charAt(0) == '0')
    {
        return 0;
    }
 
    int count0 = 1, count1 = 1, count2;
 
    // Using last two calculated values,
    // calculate for ith index
    for(int i = 2; i <= n; i++)
    {
        int dig1 = 0, dig2, dig3 = 0;
         
        // Change boolean to int
        if(digits.charAt(i - 1) != '0')
        {
            dig1 = 1;
        }
        if(digits.charAt(i - 2) == '1')
        {
            dig2 = 1;
        }
        else
            dig2 = 0;
             
        if(digits.charAt(i - 2) == '2' &&
           digits.charAt(i - 1) < '7')
        {
            dig3 = 1;
        }
        count2 = dig1 * count1 +
                 dig2 + dig3 * count0;
         
        count0 = count1;
        count1 = count2;
    }
 
    // Return the required answer
    return count1;
}
 
// Driver Code
public static void main(String[] args)
{
    String digits = "1234";
    int n = digits.length();
 
    // Function call
    System.out.print(countDecodingDP(digits, n));
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 implementation
# to count decodings
 
# A Dynamic programming based
# function to count decodings
# in digit sequence
def countDecodingDP(digits, n):
 
    # For base condition "01123"
    # should return 0
    if (digits[0] == '0'):
        return 0;   
 
    count0 = 1; count1 = 1;
 
    # Using last two calculated values,
    # calculate for ith index
    for i in range(2, n + 1):
        dig1 = 0; dig3 = 0;
 
        # Change boolean to int
        if (digits[i-1] != '0'):
            dig1 = 1;
         
        if (digits[i - 2] == '1'):
            dig2 = 1;
        else:
            dig2 = 0;
 
        if (digits[i - 2] == '2' and
            digits[i-1] < '7'):
            dig3 = 1;
         
        count2 = dig1 * count1 +
                 dig2 + dig3 * count0;
 
        count0 = count1;
        count1 = count2;   
 
    # Return the required answer
    return count1;
 
# Driver Code
if __name__ == '__main__':
    digits = "1234";
    n = len(digits);
 
    # Function call
    print(countDecodingDP(digits, n));
 
# This code is contributed by gauravrajput1

C#

// C# implementation to count decodings
using System;
 
class GFG{
 
// A Dynamic programming based function
// to count decodings in digit sequence
static int countDecodingDP(String digits, int n)
{
     
    // For base condition "01123"
    // should return 0
    if (digits[0] == '0')
    {
        return 0;
    }
 
    int count0 = 1, count1 = 1, count2;
 
    // Using last two calculated values,
    // calculate for ith index
    for(int i = 2; i <= n; i++)
    {
        int dig1 = 0, dig2, dig3 = 0;
         
        // Change bool to int
        if(digits[i - 1] != '0')
        {
            dig1 = 1;
        }
        if(digits[i - 2] == '1')
        {
            dig2 = 1;
        }
        else
            dig2 = 0;
             
        if(digits[i - 2] == '2' &&
           digits[i - 1] < '7')
        {
            dig3 = 1;
        }
        count2 = dig1 * count1 +
                 dig2 + dig3 * count0;
         
        count0 = count1;
        count1 = count2;
    }
 
    // Return the required answer
    return count1;
}
 
// Driver Code
public static void Main(String[] args)
{
    String digits = "1234";
    int n = digits.Length;
 
    // Function call
    Console.Write(countDecodingDP(digits, n));
}
}
 
// This code is contributed by Amit Katiyar

Javascript

输出：

时间复杂度： O(N)
辅助空间复杂度： O(1)
相关文章：计算给定数字序列的可能解码次数

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