给定一个字符串str和一个字符K ，任务是查找包含字符K的str的所有子字符串的计数。
例子：
Input: str = “geeks”, K = ‘g’
Output: 5
“g”, “ge”, “gee”, “geek” and “geeks” are the valid substrings.
Input: str = “geeksforgeeks”, K = ‘k’
Output: 56
天真的方法一种简单的方法是找到所有具有给定字符串的字符K的子字符串并返回计数。
高效的方法：对于字符串中的每个索引i ，找到第一个索引j ，以使i≤j且str [j] = K。现在，子字符串str [i…j]，str [i…j + 1]，str [i…j + 2]，…，str [i…n – 1]将全部包含字符K至少一次。这种方法似乎是O(N 2)在第一，但指标j不会为每个索引我再次计算，J将是所有小于j中的i值的一个有效指标。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to return the index of the
// next occurrence of character ch in str
// starting from the given index
int nextOccurrence(string str, int n,
int start, char ch)
{
for (int i = start; i < n; i++) {
// Return the index of the first
// occurrence of ch
if (str[i] == ch)
return i;
}
// No occurrence found
return -1;
}
// Function to return the count of all
// the substrings of str which contain
// the character ch at least one
int countSubStr(string str, int n, char ch)
{
// To store the count of valid substrings
int cnt = 0;
// Index of the first occurrence of ch in str
int j = nextOccurrence(str, n, 0, ch);
for (int i = 0; i < n; i++) {
while (j != -1 && j < i) {
j = nextOccurrence(str, n, j + 1, ch);
}
// No occurrence of ch after index i in str
if (j == -1)
break;
// Substrings starting at index i
// and ending at indices j, j+1, ..., n-1
// are all valid substring
cnt += (n - j);
}
return cnt;
}
// Driver code
int main()
{
string str = "geeksforgeeks";
int n = str.length();
char ch = 'k';
cout << countSubStr(str, n, ch);
return 0;
} Java
// Java implementation of the approach
import java.util.*;
class GFG
{
// Function to return the index of the
// next occurrence of character ch in str
// starting from the given index
static int nextOccurrence(String str, int n,
int start, char ch)
{
for (int i = start; i < n; i++)
{
// Return the index of the first
// occurrence of ch
if (str.charAt(i) == ch)
return i;
}
// No occurrence found
return -1;
}
// Function to return the count of all
// the substrings of str which contain
// the character ch at least one
static int countSubStr(String str,
int n, char ch)
{
// To store the count of valid substrings
int cnt = 0;
// Index of the first occurrence of ch in str
int j = nextOccurrence(str, n, 0, ch);
for (int i = 0; i < n; i++)
{
while (j != -1 && j < i)
{
j = nextOccurrence(str, n, j + 1, ch);
}
// No occurrence of ch after index i in str
if (j == -1)
break;
// Substrings starting at index i
// and ending at indices j, j+1, ..., n-1
// are all valid substring
cnt += (n - j);
}
return cnt;
}
// Driver code
static public void main ( String []arg)
{
String str = "geeksforgeeks";
int n = str.length();
char ch = 'k';
System.out.println(countSubStr(str, n, ch));
}
}
// This code is contributed by PrinciRaj1992Python3
# Python3 implementation of the approach
# Function to return the index of the
# next occurrence of character ch in strr
# starting from the given index
def nextOccurrence(strr, n, start, ch):
for i in range(start, n):
# Return the index of the first
# occurrence of ch
if (strr[i] == ch):
return i
# No occurrence found
return -1
# Function to return the count of all
# the substrrings of strr which contain
# the character ch at least one
def countSubStr(strr, n, ch):
# To store the count of valid substrrings
cnt = 0
# Index of the first occurrence of ch in strr
j = nextOccurrence(strr, n, 0, ch)
for i in range(n):
while (j != -1 and j < i):
j = nextOccurrence(strr, n, j + 1, ch)
# No occurrence of ch after index i in strr
if (j == -1):
break
# Substrrings starting at index i
# and ending at indices j, j+1, ..., n-1
# are all valid substrring
cnt += (n - j)
return cnt
# Driver code
strr = "geeksforgeeks"
n = len(strr)
ch = 'k'
print(countSubStr(strr, n, ch))
# This code is contributed by Mohit KumarC#
// C# implementation of the approach
using System;
class GFG
{
// Function to return the index of the
// next occurrence of character ch in str
// starting from the given index
static int nextOccurrence(String str, int n,
int start, char ch)
{
for (int i = start; i < n; i++)
{
// Return the index of the first
// occurrence of ch
if (str[i] == ch)
return i;
}
// No occurrence found
return -1;
}
// Function to return the count of all
// the substrings of str which contain
// the character ch at least one
static int countSubStr(String str,
int n, char ch)
{
// To store the count of valid substrings
int cnt = 0;
// Index of the first occurrence of ch in str
int j = nextOccurrence(str, n, 0, ch);
for (int i = 0; i < n; i++)
{
while (j != -1 && j < i)
{
j = nextOccurrence(str, n, j + 1, ch);
}
// No occurrence of ch after index i in str
if (j == -1)
break;
// Substrings starting at index i
// and ending at indices j, j+1, ..., n-1
// are all valid substring
cnt += (n - j);
}
return cnt;
}
// Driver code
static public void Main ()
{
String str = "geeksforgeeks";
int n = str.Length;
char ch = 'k';
Console.WriteLine(countSubStr(str, n, ch));
}
}
// This code is contributed by AnkitRai01输出：
56