字符串数组中字符串的频率
原文：https://www.geeksforgeeks.org/frequency-of-a-string-in-an-array-of-strings/
您会得到一个字符串集合和一个查询列表。 对于每个查询，都有一个字符串。 我们需要打印给定字符串在字符串集合中出现的次数。
示例：
Input : arr[] = {wer, wer, tyu, oio, tyu}
q[] = {wer, tyu, uio}
Output : 2 2 0
Explanation :
q[0] appears two times in arr[], q1[] appears
方法 1（简单）
这个想法很简单，对于每个查询字符串，我们都将其与数组中给出的所有字符串进行比较。 如果查询字符串匹配，我们将增加计数。
C++
// C++ program to find number of times a
// string appears in an array.
#include<bits/stdc++.h>
using namespace std;
// Returns count of occurrences of s in arr[]
int search(string arr[], string s, int n)
{
int counter = 0;
for(int j = 0; j < n; j++)
// Checking if string given in query
// is present in the given string.
// If present, increase times
if (s == arr[j])
counter++;
return counter;
}
void answerQueries(string arr[], string q[],
int n, int m)
{
for(int i = 0; i < m; i++)
cout << search(arr, q[i], n) << " ";
}
// Driver Code
int main()
{
string arr[] = { "aba", "baba",
"aba", "xzxb" };
string q[] = { "aba", "xzxb", "ab" };
int n = sizeof(arr) / sizeof(arr[0]);
int m = sizeof(q) / sizeof(q[0]);
answerQueries(arr, q, n, m);
}
// This code is contributed by rutvik_56
Java
// Java program to find number of times a string
// appears in an array.
class SubString
{
/* Returns count of occurrences of s in arr[] */
static int search(String[]arr, String s)
{
int counter = 0;
for (int j = 0; j < arr.length; j++)
/* checking if string given in query is
present in the given string. If present,
increase times*/
if (s.equals(arr[j]))
counter++;
return counter;
}
static void answerQueries(String[] arr, String q[])
{
for (int i=0;i<q.length; i++)
System.out.print(search(arr, q[i]) + " ");
}
/* driver code*/
public static void main(String[] args) {
String[] arr = {"aba","baba","aba","xzxb"};
String[] q = {"aba","xzxb","ab"};
answerQueries(arr, q);
}
}
Python3
# Python3 program to find number of
# times a string appears in an array.
# Returns count of occurrences of s in arr[]
def search(arr, s):
counter = 0
for j in range(len(arr)):
# checking if string given in query
# is present in the given string.
# If present, increase times
if (s == (arr[j])):
counter += 1
return counter
def answerQueries(arr, q):
for i in range(len(q)):
print(search(arr, q[i]),
end = " ")
# Driver code
if __name__ == '__main__':
arr = ["aba", "baba", "aba", "xzxb"]
q = ["aba", "xzxb", "ab"]
answerQueries(arr, q)
# This code is contributed
# by PrinciRaj19992
C
// C# program to find number of
// times a string appears in an array.
using System;
class SubString
{
/* Returns count of occurrences of s in arr[] */
static int search(String[]arr, String s)
{
int counter = 0;
for (int j = 0; j < arr.Length; j++)
/* checking if string given in query is
present in the given string. If present,
increase times*/
if (s.Equals(arr[j]))
counter++;
return counter;
}
static void answerQueries(String []arr, String []q)
{
for (int i = 0; i < q.Length; i++)
Console.Write(search(arr, q[i]) + " ");
}
// Driver code
public static void Main()
{
String []arr = {"aba","baba","aba","xzxb"};
String []q = {"aba","xzxb","ab"};
answerQueries(arr, q);
}
}
//This code is contributed by nitin mittal
PHP
<?php
# Php program to find number of
# times a string appears in an array.
# Returns count of occurrences of s in arr[]
function search($arr, $s)
{
$counter = 0;
for ($j=0; $j<count($arr); $j++)
{
# checking if string given in query
# is present in the given string.
# If present, increase times
if ($s == ($arr[$j]))
$counter += 1;
}
return $counter;
}
function answerQueries($arr, $q)
{
for ($i=0; $i<count($q); $i++)
{
echo(search($arr, $q[$i]));
echo (" ");
}
}
# Driver code
$arr = array("aba", "baba", "aba", "xzxb");
$q = array("aba", "xzxb", "ab");
answerQueries($arr, $q);
# This code is contributed
# by Shivi_Aggarwal
?>
输出：
2 1 0
方法 2（使用 Trie）
Trie 一种有效的数据结构，用于增强和检索类似字符串的数据。 搜索复杂度是最佳的键长度。
在此解决方案中，我们将字符串集合插入到 Trie 数据结构中，以便将它们存储在其中。 一件重要的事情是，我们要统计叶节点中的出现次数。 然后，在 Trie 中搜索给定的查询字符串，并检查 Trie 中是否存在该字符串。
CPP
// C++ program to count number of times
// a string appears in an array of strings
#include<iostream>
using namespace std;
const int MAX_CHAR = 26;
struct Trie
{
// To store number of times
// a string is present. It is
// 0 is string is not present
int cnt;
Trie *node[MAX_CHAR];
Trie()
{
for(int i=0; i<MAX_CHAR; i++)
node[i] = NULL;
cnt = 0;
}
};
/* function to insert a string into the Trie*/
Trie *insert(Trie *root,string s)
{
Trie *temp = root;
int n = s.size();
for(int i=0; i<n; i++)
{
/*calculation ascii value*/
int index = s[i]-'a';
/* If the given node is not already
present in the Trie than create
the new node*/
if (!root->node[index])
root->node[index] = new Trie();
root = root->node[index];
}
root->cnt++;
return temp;
}
/* Returns count of occurrences of s in Trie*/
int search(Trie *root, string s)
{
int n = s.size();
for (int i=0; i<n; i++)
{
int index = s[i]-'a';
if (!root->node[index])
return 0;
root = root->node[index];
}
return root->cnt;
}
void answerQueries(string arr[], int n, string q[],
int m)
{
Trie *root = new Trie();
/* inserting in Trie */
for (int i=0; i<n; i++)
root = insert(root,arr[i]);
/* searching the strings in Trie */
for (int i=0; i<m; i++)
cout << search(root, q[i]) << " ";
}
/* Driver code */
int main()
{
string arr[] = {"aba","baba","aba","xzxb"};
int n = sizeof(arr)/sizeof(arr[0]);
string q[] = {"aba","xzxb","ab"};
int m = sizeof(q)/sizeof(q[0]);
answerQueries(arr, n, q, m);
return 0;
}
输出：
2
1
0
方法 3（哈希）
我们可以使用哈希映射并将所有给定的字符串插入其中。 对于每个查询字符串，我们只需要在哈希映射中进行查找即可。
请参考字典数据结构，以比较基于哈希和基于 Trie 的解决方案。
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