找出字符可以重新排列形成给定单词的子串的个数
原文:https://www . geeksforgeeks . org/find-子字符串的计数-其字符可以重新排列以形成给定的单词/
给定一个字符串 str ，任务是找到所有长度为 4 的子字符串的计数，这些子字符串的字符可以重新排列以形成单词“拍手”。 例:
输入:str = " clapc " T3】输出:2 “clap”和“lapc”是所需的子串 输入: str = "abcd" 输出: 0
方法:对于每一个长度为 4 的子串，从单词【拍拍】开始计算字符出现的次数。如果每个字符在子字符串中都恰好出现一次，则增加计数。最后打印计数。 以下是上述方法的实施:
C++
// CPP implementation of the approach
#include<bits/stdc++.h>
using namespace std;
// Function to return the count of
// required occurrence
int countOcc(string s)
{
// To store the count of occurrences
int cnt = 0;
// Check first four characters from ith position
for (int i = 0; i < s.length() - 3; i++)
{
// Variables for counting the required characters
int c = 0, l = 0, a = 0, p = 0;
// Check the four contiguous characters which
// can be reordered to form 'clap'
for (int j = i; j < i + 4; j++)
{
switch (s[j])
{
case 'c':
c++;
break;
case 'l':
l++;
break;
case 'a':
a++;
break;
case 'p':
p++;
break;
}
}
// If all four contiguous characters are present
// then increment cnt variable
if (c == 1 && l == 1 && a == 1 && p == 1)
cnt++;
}
return cnt;
}
// Driver code
int main()
{
string s = "clapc";
transform(s.begin(), s.end(), s.begin(), ::tolower);
cout << (countOcc(s));
}
// This code is contributed by
// Surendra_Gangwar
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
class GFG {
// Function to return the count of
// required occurrence
static int countOcc(String s)
{
// To store the count of occurrences
int cnt = 0;
// Check first four characters from ith position
for (int i = 0; i < s.length() - 3; i++) {
// Variables for counting the required characters
int c = 0, l = 0, a = 0, p = 0;
// Check the four contiguous characters which
// can be reordered to form 'clap'
for (int j = i; j < i + 4; j++) {
switch (s.charAt(j)) {
case 'c':
c++;
break;
case 'l':
l++;
break;
case 'a':
a++;
break;
case 'p':
p++;
break;
}
}
// If all four contiguous characters are present
// then increment cnt variable
if (c == 1 && l == 1 && a == 1 && p == 1)
cnt++;
}
return cnt;
}
// Driver code
public static void main(String args[])
{
String s = "clapc";
System.out.print(countOcc(s.toLowerCase()));
}
}
Python 3
# Python3 implementation of the approach
# Function to return the count of
# required occurrence
def countOcc(s):
# To store the count of occurrences
cnt = 0
# Check first four characters from ith position
for i in range(0, len(s) - 3):
# Variables for counting the required characters
c, l, a, p = 0, 0, 0, 0
# Check the four contiguous characters
# which can be reordered to form 'clap'
for j in range(i, i + 4):
if s[j] == 'c':
c += 1
elif s[j] == 'l':
l += 1
elif s[j] == 'a':
a += 1
elif s[j] == 'p':
p += 1
# If all four contiguous characters are
# present then increment cnt variable
if c == 1 and l == 1 and a == 1 and p == 1:
cnt += 1
return cnt
# Driver code
if __name__ == "__main__":
s = "clapc"
print(countOcc(s.lower()))
# This code is contributed by Rituraj Jain
C
// C# implementation of the approach
using System;
class GFG
{
// Function to return the count of
// required occurrence
static int countOcc(string s)
{
// To store the count of occurrences
int cnt = 0;
// Check first four characters
// from ith position
for (int i = 0; i < s.Length - 3; i++)
{
// Variables for counting the
// required characters
int c = 0, l = 0, a = 0, p = 0;
// Check the four contiguous characters
// which can be reordered to form 'clap'
for (int j = i; j < i + 4; j++)
{
switch (s[j])
{
case 'c':
c++;
break;
case 'l':
l++;
break;
case 'a':
a++;
break;
case 'p':
p++;
break;
}
}
// If all four contiguous characters are
// present then increment cnt variable
if (c == 1 && l == 1 && a == 1 && p == 1)
cnt++;
}
return cnt;
}
// Driver code
public static void Main()
{
string s = "clapc";
Console.Write(countOcc(s.ToLower()));
}
}
// This code is contributed by Akanksha Rai
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation of the approach
// Function to return the count of
// required occurrence
function countOcc($s)
{
// To store the count of occurrences
$cnt = 0;
// Check first four characters
// from ith position
for ($i = 0; $i < strlen($s) - 3; $i++)
{
// Variables for counting the
// required characters
$c = 0; $l = 0; $a = 0; $p = 0;
// Check the four contiguous characters
// which can be reordered to form 'clap'
for ($j = $i; $j < $i + 4; $j++)
{
switch ($s[$j])
{
case 'c':
$c++;
break;
case 'l':
$l++;
break;
case 'a':
$a++;
break;
case 'p':
$p++;
break;
}
}
// If all four contiguous characters are present
// then increment cnt variable
if ($c == 1 && $l == 1 &&
$a == 1 && $p == 1)
$cnt++;
}
return $cnt;
}
// Driver code
$s = "clapc";
echo countOcc(strtolower($s));
// This code is contributed by Ryuga
?>
java 描述语言
<script>
// Javascript implementation of the approach
// Function to return the count of
// required occurrence
function countOcc(s)
{
// To store the count of occurrences
var cnt = 0;
// Check first four characters from ith position
for (var i = 0; i < s.length - 3; i++)
{
// Variables for counting the required characters
var c = 0, l = 0, a = 0, p = 0;
// Check the four contiguous characters which
// can be reordered to form 'clap'
for (var j = i; j < i + 4; j++)
{
switch (s[j])
{
case 'c':
c++;
break;
case 'l':
l++;
break;
case 'a':
a++;
break;
case 'p':
p++;
break;
}
}
// If all four contiguous characters are present
// then increment cnt variable
if (c == 1 && l == 1 && a == 1 && p == 1)
cnt++;
}
return cnt;
}
// Driver code
var s = "clapc";
s = s.toLowerCase();
document.write(countOcc(s));
</script>
Output:
2
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