给定三个分别由N 、 A和B字符组成的字符串S 、 X和Y ，任务是在给定的字符串S 中每次出现子串Y之前找到子串X出现的次数。
例子：
Input S = ”abcdefdefabc”, X = ”def”, Y = ”abc”
Output: 0 2
Explanation:
First occurence of Y(= “abc”): No of occurrences of X(= “def”) = 0.
Second occurrence of Y: No of occurrences of X = 0.
Input: S = ”accccbbbbbbaaa”, X = ”a”, Y = ”b”
Output: 0 6 6 6
处理方法：按照以下步骤解决问题：
- 初始化一个变量，比如count ，它存储X的总出现次数。
- 遍历给定的字符串S并执行以下步骤：
- 如果[i, B]范围内的子字符串等于Y ，则将count增加1 。
- 如果子串在范围[I，A]是等于X，然后打印计数的值作为字符串Y的字符串X的当前发生之前得到的计数。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to count occurrences of
// the string Y in the string S for
// every occurrence of X in S
void countOccurrences(string S,
string X,
string Y)
{
// Stores the count of
// occurrences of X
int count = 0;
// Stores the lengths of the
// three strings
int N = S.length(), A = X.length();
int B = Y.length();
// Traverse the string S
for (int i = 0; i < N; i++) {
// If the current substring
// is Y, then increment the
// value of count by 1
if (S.substr(i, B) == Y)
count++;
// If the current substring
// is X, then print the count
if (S.substr(i, A) == X)
cout << count << " ";
}
}
// Driver Code
int main()
{
string S = "abcdefdefabc";
string X = "abc";
string Y = "def";
countOccurrences(S, X, Y);
return 0;
} Java
// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
public class GFG {
// Function to count occurrences of
// the string Y in the string S for
// every occurrence of X in S
static void countOccurrences(String S, String X,
String Y)
{
// Stores the count of
// occurrences of X
int count = 0;
// Stores the lengths of the
// three strings
int N = S.length(), A = X.length();
int B = Y.length();
// Traverse the string S
for (int i = 0; i < N; i++) {
// If the current substring
// is Y, then increment the
// value of count by 1
if (S.substring(i, Math.min(N, i + B))
.equals(Y))
count++;
// If the current substring
// is X, then print the count
if (S.substring(i, Math.min(N, i + A))
.equals(X))
System.out.print(count + " ");
}
}
// Driver Code
public static void main(String[] args)
{
String S = "abcdefdefabc";
String X = "abc";
String Y = "def";
countOccurrences(S, X, Y);
}
}
// This code is contributed by Kingash.Python3
# Python program for the above approach
# Function to count occurrences of
# the string Y in the string S for
# every occurrence of X in S
def countOccurrences(S, X, Y):
# Stores the count of
# occurrences of X
count = 0
# Stores the lengths of the
# three strings
N = len(S)
A = len(X)
B = len(Y)
# Traverse the string S
for i in range( 0, N):
# If the current substring
# is Y, then increment the
# value of count by 1
if (S[i: i+B] == Y):
count+=1
# If the current substring
# is X, then print the count
if (S[i:i+A] == X):
print(count, end = " ")
# Driver Code
S = "abcdefdefabc"
X = "abc"
Y = "def"
countOccurrences(S, X, Y)
# This code is contributed by rohitsingh07052.C#
// C# program for the above approach
using System;
public class GFG {
// Function to count occurrences of
// the string Y in the string S for
// every occurrence of X in S
static void countOccurrences(string S, string X,
string Y)
{
// Stores the count of
// occurrences of X
int count = 0;
// Stores the lengths of the
// three strings
int N = S.Length, A = X.Length;
int B = Y.Length;
int P = Math.Min(A, Math.Min(N, B));
// Traverse the string S
for (int i = 0; i < N - P + 1; i++) {
// If the current substring
// is Y, then increment the
// value of count by 1
if (S.Substring(i, Math.Min(N, B)).Equals(Y))
count++;
// If the current substring
// is X, then print the count
if (S.Substring(i, Math.Min(N, A)).Equals(X))
Console.Write(count + " ");
}
}
// Driver Code
public static void Main(string[] args)
{
string S = "abcdefdefabc";
string X = "abc";
string Y = "def";
countOccurrences(S, X, Y);
}
}
// This code is contributed by ukasp.Javascript
输出：
0 2时间复杂度： O(N*(A + B))
辅助空间： O(1)
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