假设数字为n，则计算长度为n的字符串数，以使每个字符串具有相邻字符，且ASCII值之差为1。
例子：
Input : N = 1
Output : Total strings are 26
Explanation : For N=1, strings
are a, b, c,, ...., x, y, z
Input : N = 2
Output : Total strings are 50
Explanation : For N = 2, strings
are ab, ba, bc, cb, .., yx, yz, zy
对于开始字符“A”和长度“我”的字符串，我们认为长的“I-1”的所有字符串，并开始用字符“B”
为了启动用字符“G”和长度“I”的字符串，我们考虑长度“I-1”的所有字符串，并具有字符“H”开始，并用“F”长度“I-1”和开始的所有字符串。
我们取基础案例对于n = 1，和设定结果为所有26字符为1。这仅仅意味着当1个是考虑从AZ采取只有一旦所有字母。
对于N = 2 ， 
对于N = 3 ， 
结论：对于N = n 
countAdjacent(n)
dp[i][j] finally stores count of strings
of length i and starting with
character j.
Initialize dp[n+1][27] as 0
Initialize dp[1][j] = 1 where j = 0 to 25
for i = 2 to n
for j = 0 to 25
if (j = 0)
dp[i][j] = dp[i-1][j+1];
else
dp[i][j] = dp[i-1][j-1] + dp[i-1][j+1];
Sum of n-th row from 0 to 25 is the result.
C++
// CPP Program to count strings with adjacent
// characters.
#include
using namespace std;
int countStrs(int n)
{
long int dp[n + 1][27];
// Initializing arr[n+1][27] to 0
memset(dp, 0, sizeof(dp));
// Initialing 1st row all 1 from 0 to 25
for (int i = 0; i <= 25; i++)
dp[1][i] = 1;
// Begin evaluating from i=2 since 1st row is set
for (int i = 2; i <= n; i++) {
for (int j = 0; j <= 25; j++)
// j=0 is 'A' which can make strings
// of length i using strings of length
// i-1 and starting with 'B'
if (j == 0)
dp[i][j] = dp[i - 1][j + 1];
else
dp[i][j] = (dp[i - 1][j - 1] +
dp[i - 1][j + 1]);
}
// Our result is sum of last row.
long int sum = 0;
for (int i = 0; i <= 25; i++)
sum = (sum + dp[n][i]);
return sum;
}
// Driver's Code
int main()
{
int n = 3;
cout << "Total strings are : " << countStrs(n);
return 0;
} Java
// Java Program to count strings
// with adjacent characters.
class GFG
{
static long countStrs(int n)
{
long[][] dp = new long[n + 1][27];
// Initializing arr[n+1][27] to 0
for (int i = 0; i < n + 1; i++)
{
for (int j = 0; j < 27; j++)
{
dp[i][j] = 0;
}
}
// Initialing 1st row all 1 from 0 to 25
for (int i = 0; i <= 25; i++)
{
dp[1][i] = 1;
}
// Begin evaluating from i=2
// since 1st row is set
for (int i = 2; i <= n; i++)
{
// j=0 is 'A' which can make strings
for (int j = 0; j <= 25; j++)
// of length i using strings of length
// i-1 and starting with 'B'
{
if (j == 0)
{
dp[i][j] = dp[i - 1][j + 1];
}
else
{
dp[i][j] = (dp[i - 1][j - 1]
+ dp[i - 1][j + 1]);
}
}
}
// Our result is sum of last row.
long sum = 0;
for (int i = 0; i <= 25; i++)
{
sum = (sum + dp[n][i]);
}
return sum;
}
// Driver Code
public static void main(String[] args)
{
int n = 3;
System.out.println("Total strings are : " +
countStrs(n));
}
}
// This code is contributed by 29AjayKumarPython 3
# Python3 Program to count strings with
# adjacent characters.
def countStrs(n):
# Initializing arr[n+1][27] to 0
dp = [[0 for j in range(27)]
for i in range(n + 1)]
# Initialing 1st row all 1 from 0 to 25
for i in range(0, 26):
dp[1][i] = 1
# Begin evaluating from i=2 since
# 1st row is set
for i in range(2, n + 1):
for j in range(0, 26):
# j=0 is 'A' which can make strings
# of length i using strings of length
# i-1 and starting with 'B'
if(j == 0):
dp[i][j] = dp[i - 1][j + 1];
else:
dp[i][j] = (dp[i - 1][j - 1] +
dp[i - 1][j + 1])
# Our result is sum of last row.
sum = 0
for i in range(0, 26):
sum = sum + dp[n][i]
return sum
# Driver's Code
if __name__ == "__main__":
n = 3
print("Total strings are : ", countStrs(n))
# This code is contributed by Sairahul JellaC#
// C# Program to count strings with
// adjacent characters.
using System;
class GFG
{
static long countStrs(int n)
{
long[,] dp = new long[n + 1, 27];
// Initializing arr[n+1][27] to 0
for(int i = 0; i < n + 1; i++)
for(int j = 0; j < 27; j++)
dp[i, j] = 0;
// Initialing 1st row all 1 from 0 to 25
for (int i = 0; i <= 25; i++)
dp[1, i] = 1;
// Begin evaluating from i=2 since 1st row is set
for (int i = 2; i <= n; i++)
{
for (int j = 0; j <= 25; j++)
// j=0 is 'A' which can make strings
// of length i using strings of length
// i-1 and starting with 'B'
if (j == 0)
dp[i, j] = dp[i - 1, j + 1];
else
dp[i, j] = (dp[i - 1, j - 1] +
dp[i - 1, j + 1]);
}
// Our result is sum of last row.
long sum = 0;
for (int i = 0; i <= 25; i++)
sum = (sum + dp[n, i]);
return sum;
}
// Driver Code
static void Main()
{
int n = 3;
Console.Write("Total strings are : " + countStrs(n));
}
}
// This code is contributed by DrRoot_输出：
Total strings are : 98