存在于其他
中的一个字符串的子字符串数量
原文:https://www . geeksforgeeks . org/one-string-present-in-other 的子字符串数/
假设给我们一个字符串 s1，我们需要找到字符串 s2 中出现的 s1 的子字符串(包括同一子字符串的多次出现)的总数。
示例:
Input : s1 = aab
s2 = aaaab
Output :6
Substrings of s1 are ["a", "a", "b", "aa",
"ab", "aab"]. These all are present in s2\.
Hence, answer is 6.
Input :s1 = abcd
s2 = swalencud
Output :3
想法是考虑 s1 的所有子串，并检查它是否出现在 s2 中。
C++
// CPP program to count number of substrings of s1
// present in s2.
#include<iostream>
#include<string>
using namespace std;
int countSubstrs(string s1, string s2)
{
int ans = 0;
for (int i = 0; i < s1.length(); i++) {
// s3 stores all substrings of s1
string s3;
for (int j = i; j < s1.length(); j++) {
s3 += s1[j];
// check the presence of s3 in s2
if (s2.find(s3) != string::npos)
ans++;
}
}
return ans;
}
// Driver code
int main()
{
string s1 = "aab", s2 = "aaaab";
cout << countSubstrs(s1, s2);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to count number of
// substrings of s1 present in s2.
import java.util.*;
class GFG
{
static int countSubstrs(String s1,
String s2)
{
int ans = 0;
for (int i = 0; i < s1.length(); i++)
{
// s3 stores all substrings of s1
String s3 = "";
char[] s4 = s1.toCharArray();
for (int j = i; j < s1.length(); j++)
{
s3 += s4[j];
// check the presence of s3 in s2
if (s2.indexOf(s3) != -1)
ans++;
}
}
return ans;
}
// Driver code
public static void main(String[] args)
{
String s1 = "aab", s2 = "aaaab";
System.out.println(countSubstrs(s1, s2));
}
}
// This code is contributed by ChitraNayal
Python 3
# Python 3 program to count number of substrings of s1
# present in s2.
# Function for counting no. of substring
# of s1 present in s2
def countSubstrs(s1, s2) :
ans = 0
for i in range(len(s1)) :
s3 = ""
# s3 stores all substrings of s1
for j in range(i, len(s1)) :
s3 += s1[j]
# check the presence of s3 in s2
if s2.find(s3) != -1 :
ans += 1
return ans
# Driver code
if __name__ == "__main__" :
s1 = "aab"
s2 = "aaaab"
# function calling
print(countSubstrs(s1, s2))
# This code is contributed by ANKITRAI1
C
// C# program to count number of
// substrings of s1 present in s2.
using System;
class GFG
{
static int countSubstrs(String s1,
String s2)
{
int ans = 0;
for (int i = 0; i < s1.Length; i++)
{
// s3 stores all substrings of s1
String s3 = "";
char[] s4 = s1.ToCharArray();
for (int j = i; j < s1.Length; j++)
{
s3 += s4[j];
// check the presence of s3 in s2
if (s2.IndexOf(s3) != -1)
ans++;
}
}
return ans;
}
// Driver code
public static void Main(String[] args)
{
String s1 = "aab", s2 = "aaaab";
Console.WriteLine(countSubstrs(s1, s2));
}
}
// This code is contributed
// by Kirti_Mangal
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to count number of
// substrings of s1 present in s2.
function countSubstrs($s1, $s2)
{
$ans = 0;
for ($i = 0; $i < strlen($s1); $i++)
{
// s3 stores all substrings of s1
$s3 = "";
for ($j = $i;
$j < strlen($s1); $j++)
{
$s3 += $s1[$j];
// check the presence of s3 in s2
if (stripos($s2, $s3, 0) != -1)
$ans++;
}
}
return $ans;
}
// Driver code
$s1 = "aab";
$s2 = "aaaab";
echo countSubstrs($s1, $s2);
// This code is contributed
// by ChitraNayal
?>
java 描述语言
<script>
// javascript program to count number of
// substrings of s1 present in s2.
function countSubstrs( s1, s2)
{
var ans = 0;
for (var i = 0; i < s1.length; i++)
{
// s3 stores all substrings of s1
var s3 = "";
var s4 = s1 ;
for (var j = i; j < s1.length; j++)
{
s3 += s4[j];
// check the presence of s3 in s2
if (s2.indexOf(s3) != -1)
ans++;
}
}
return ans;
}
// Driver code
var s1 = "aab", s2 = "aaaab";
document.write(countSubstrs(s1, s2));
</script>
Output:
6
版权属于：月萌API www.moonapi.com，转载请注明出处
