给定字符串“ s”和字符“ c”，任务是查找字符串的排列数，其中所有出现的字符“ c”将在一起(一个接一个)。
例子 ：
Input: Str = “AKA” ch = ‘A’
Output: 2
All the unique permutations of AKA are : AKA, AAK and KAA
‘A’ comes consecutively only twice. Hence, the answer is 2
Input: str= “MISSISSIPPI” ch = ‘S’
Output: 840
天真的方法：
- 生成给定字符串的所有排列。
- 对于每个排列，检查是否所有出现的字符都一起出现。
- 上一步的排列数就是答案。
高效方法：将字符串的长度设为“ l”，并将字符串字符的频率设为“ n”。
- 存储字符串中每个字符的频率。
- 如果字符中不存在该字符串，则输出将为“ 0”。
- 将所有出现的字符视为一个组合的单个字符。
- 因此，“ l”将变为“ l – occ + 1”，其中“ occ”是给定字符的总出现次数，“ l”是字符串的新长度。
- 返回((l的因子)/(给定字符以外的每个字符的出现次数的阶乘积的乘积))
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to return factorial
// of the number passed as argument
long long int fact(int n)
{
long long result = 1;
for (int i = 1; i <= n; i++)
result *= i;
return result;
}
// Function to get the total permutations
// which satisfy the given condition
int getResult(string str, char ch)
{
// Create has to store count
// of each character
int has[26] = { 0 };
// Store character occurrences
for (int i = 0; i < str.length(); i++)
has[str[i] - 'A']++;
// Count number of times
// Particular character comes
int particular = has[ch - 'A'];
// If particular character isn't
// present in the string then return 0
if (particular == 0)
return 0;
// Remove count of particular character
has[ch - 'A'] = 0;
// Total length
// of the string
int total = str.length();
// Assume all occurrences of
// particular character as a
// single character.
total = total - particular + 1;
// Compute factorial of the length
long long int result = fact(total);
// Divide by the factorials of
// the no. of occurrences of all
// the characters.
for (int i = 0; i < 26; i++) {
if (has[i] > 1) {
result = result / fact(has[i]);
}
}
// return the result
return result;
}
// Driver Code
int main()
{
string str = "MISSISSIPPI";
// Assuming the string and the character
// are all in uppercase
cout << getResult(str, 'S') << endl;
return 0;
} Java
// Java implementation of above approach
import java.util.*;
class solution
{
// Function to return factorial
// of the number passed as argument
static int fact(int n)
{
int result = 1;
for (int i = 1; i <= n; i++)
result *= i;
return result;
}
// Function to get the total permutations
// which satisfy the given condition
static int getResult(String str, char ch)
{
// Create has to store count
// of each character
int has[] = new int[26];
for(int i=0;i<26;i++)
has[i]=0;
// Store character occurrences
for (int i = 0; i < str.length(); i++)
has[str.charAt(i) - 'A']++;
// Count number of times
// Particular character comes
int particular = has[ch - 'A'];
// If particular character isn't
// present in the string then return 0
if (particular == 0)
return 0;
// Remove count of particular character
has[ch - 'A'] = 0;
// Total length
// of the string
int total = str.length();
// Assume all occurrences of
// particular character as a
// single character.
total = total - particular + 1;
// Compute factorial of the length
int result = fact(total);
// Divide by the factorials of
// the no. of occurrences of all
// the characters.
for (int i = 0; i < 26; i++) {
if (has[i] > 1) {
result = result / fact(has[i]);
}
}
// return the result
return result;
}
// Driver Code
public static void main(String args[])
{
String str = "MISSISSIPPI";
// Assuming the string and the character
// are all in uppercase
System.out.println( getResult(str, 'S') );
}
}
//contributed by Arnab KunduPython 3
# Python3 implementation of
# the approach
# Function to return factorial
# of the number passed as argument
def fact(n) :
result = 1
for i in range(1, n + 1) :
result *= i
return result
# Function to get the total permutations
# which satisfy the given condition
def getResult(string, ch):
# Create has to store count
# of each character
has = [0] * 26
# Store character occurrences
for i in range(len(string)) :
has[ord(string[i]) - ord('A')] += 1
# Count number of times
# Particular character comes
particular = has[ord(ch) - ord('A')]
# If particular character isn't
# present in the string then return 0
if particular == 0 :
return 0
# Remove count of particular character
has[ord(ch) - ord('A')] = 0
# Total length
# of the string
total = len(string)
# Assume all occurrences of
# particular character as a
# single character.
total = total - particular + 1
# Compute factorial of the length
result = fact(total)
# Divide by the factorials of
# the no. of occurrences of all
# the characters.
for i in range(26) :
if has[i] > 1 :
result /= fact(has[i])
# return the result
return result
# Driver code
if __name__ == "__main__" :
string = "MISSISSIPPI"
# Assuming the string and the character
# are all in uppercase
print(getResult(string,'S'))
# This code is contributed
# by ANKITRAI1C#
// C# implementation of above approach
using System;
class GFG
{
// Function to return factorial
// of the number passed as argument
static int fact(int n)
{
int result = 1;
for (int i = 1; i <= n; i++)
result *= i;
return result;
}
// Function to get the total permutations
// which satisfy the given condition
static int getResult(string str, char ch)
{
// Create has to store count
// of each character
int []has = new int[26];
for(int i = 0; i < 26; i++)
has[i] = 0;
// Store character occurrences
for (int i = 0; i < str.Length; i++)
has[str[i] - 'A']++;
// Count number of times
// Particular character comes
int particular = has[ch - 'A'];
// If particular character
// isn't present in the string
// then return 0
if (particular == 0)
return 0;
// Remove count of particular character
has[ch - 'A'] = 0;
// Total length of the string
int total = str.Length;
// Assume all occurrences of
// particular character as a
// single character.
total = total - particular + 1;
// Compute factorial of the length
int result = fact(total);
// Divide by the factorials of
// the no. of occurrences of all
// the characters.
for (int i = 0; i < 26; i++)
{
if (has[i] > 1)
{
result = result / fact(has[i]);
}
}
// return the result
return result;
}
// Driver Code
public static void Main()
{
string str = "MISSISSIPPI";
// Assuming the string and the
// character are all in uppercase
Console.WriteLine(getResult(str, 'S') );
}
}
// This code is contributed by anuj_67PHP
1)
{
$result = $result / fact($has[$i]);
}
}
// return the result
return $result;
}
// Driver Code
$str = "MISSISSIPPI";
// Assuming the string and the character
// are all in uppercase
echo getResult($str, 'S')."\n" ;
// This code is contributed by ita_c
?>输出：
840