字符串中子串的频率
原文:https://www.geeksforgeeks.org/frequency-substring-string/
给定输入字符串和子字符串。查找给定字符串中子字符串出现的频率。
例:
Input : man (pattern)
dhimanman (string)
Output : 2
Input : nn (pattern)
Banana (String)
Output : 0
Input : man (pattern)
dhimanman (string)
Output : 2
Input : aa (pattern)
aaaaa (String)
Output : 4
一个简单的解决方法就是一个一个的匹配字符。每当我们看到完全匹配时，我们就增加计数。以下是基于朴素模式搜索的简单解决方案。
c++
// Simple C++ program to count occurrences
// of pat in txt.
#include<bits/stdc++.h>
using namespace std;
int countFreq(string &pat, string &txt)
{
int M = pat.length();
int N = txt.length();
int res = 0;
/* A loop to slide pat[] one by one */
for (int i = 0; i <= N - M; i++)
{
/* For current index i, check for
pattern match */
int j;
for (j = 0; j < M; j++)
if (txt[i+j] != pat[j])
break;
// if pat[0...M-1] = txt[i, i+1, ...i+M-1]
if (j == M)
{
res++;
}
}
return res;
}
/* Driver program to test above function */
int main()
{
string txt = "dhimanman";
string pat = "man";
cout << countFreq(pat, txt);
return 0;
}
Java
// Simple Java program to count occurrences
// of pat in txt.
class GFG {
static int countFreq(String pat, String txt) {
int M = pat.length();
int N = txt.length();
int res = 0;
/* A loop to slide pat[] one by one */
for (int i = 0; i <= N - M; i++) {
/* For current index i, check for
pattern match */
int j;
for (j = 0; j < M; j++) {
if (txt.charAt(i + j) != pat.charAt(j)) {
break;
}
}
// if pat[0...M-1] = txt[i, i+1, ...i+M-1]
if (j == M) {
res++;
j = 0;
}
}
return res;
}
/* Driver program to test above function */
static public void main(String[] args) {
String txt = "dhimanman";
String pat = "man";
System.out.println(countFreq(pat, txt));
}
}
// This code is contributed by 29AjayKumar
python 3
# Simple python program to count
# occurrences of pat in txt.
def countFreq(pat, txt):
M = len(pat)
N = len(txt)
res = 0
# A loop to slide pat[] one by one
for i in range(N - M + 1):
# For current index i, check
# for pattern match
j = 0
while j < M:
if (txt[i + j] != pat[j]):
break
j += 1
if (j == M):
res += 1
j = 0
return res
# Driver Code
if __name__ == '__main__':
txt = "dhimanman"
pat = "man"
print(countFreq(pat, txt))
# This code is contributed
# by PrinciRaj1992
c
// Simple C# program to count occurrences
// of pat in txt.
using System;
public class GFG {
static int countFreq(String pat, String txt) {
int M = pat.Length;
int N = txt.Length;
int res = 0;
/* A loop to slide pat[] one by one */
for (int i = 0; i <= N - M; i++) {
/* For current index i, check for
pattern match */
int j;
for (j = 0; j < M; j++) {
if (txt[i + j] != pat[j]) {
break;
}
}
// if pat[0...M-1] = txt[i, i+1, ...i+M-1]
if (j == M) {
res++;
j = 0;
}
}
return res;
}
/* Driver program to test above function */
static public void Main() {
String txt = "dhimanman";
String pat = "man";
Console.Write(countFreq(pat, txt));
}
}
// This code is contributed by 29AjayKumar
PHP
<?php
// Simple PHP program to count occurrences
// of pat in txt.
function countFreq($pat, $txt)
{
$M = strlen($pat);
$N = strlen($txt);
$res = 0;
/* A loop to slide pat[] one by one */
for ($i = 0; $i <= $N - $M; $i++)
{
/* For current index i, check for
pattern match */
for ($j = 0; $j < $M; $j++)
if ($txt[$i+$j] != $pat[$j])
break;
// if pat[0...M-1] = txt[i, i+1, ...i+M-1]
if ($j == $M)
{
$res++;
$j = 0;
}
}
return $res;
}
// Driver Code
$txt = "dhimanman";
$pat = "man";
echo countFreq($pat, $txt);
// This code is contributed
// by Akanksha Rai
Javascript
<script>
// JavaScript program to count occurrences
// of pat in txt.
let mod = 100000007;
function countFreq(pat, txt)
{
let M = pat.length;
let N = txt.length;
let res = 0;
// A loop to slide pat[] one by one
for(let i = 0; i <= N - M; i++)
{
// For current index i, check for
// pattern match
let j;
for(j = 0; j < M; j++)
{
if (txt[i + j] != pat[j])
{
break;
}
}
// If pat[0...M-1] = txt[i, i+1, ...i+M-1]
if (j == M)
{
res++;
j = 0;
}
}
return res;
}
// Driver Code
let txt = "dhimanman";
let pat = "man";
document.write(countFreq(pat, txt));
// This code is contributed by code_hunt
</script>
输出
2
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