给定一个数字序列S ,任务是查找给定数字序列的可能解码数目,其中1代表“ A”,2代表“ B”……依此类推,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:为了解决O(N)时间复杂度的问题,使用了动态编程。为了将辅助空间的复杂度降低到O(1),我们使用了Fibonacci Number Post中讨论的递归关系的空间优化版本。
类似于斐波纳契数,可以使用前两个索引来计算任何当前“ i th ”索引的关键观测值。因此,用于计算第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
输出:
3
时间复杂度: O(N)
辅助空间复杂度: O(1)
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