计数并打印 ASCII 值不在[l，r]
范围内的字母
原文:https://www . geesforgeks . org/count-and-print-the-alphabets-having-ascii-value-not-in-range-l-r/
给定一个字符串 str ，任务是计算具有 ASCII 值的字母数量，不在[l，r]范围内。 例:
Input: str = "geeksforgeeks", l = 102, r = 111
Output: Count = 7
Characters - e, s, r have ascii values not in the range [102, 111].
Input: str = "GeEkS", l = 80, r = 111
Output: Count = 2
方法:开始遍历字符串，检查当前字符的 ASCII 值是否小于等于 r 且大于等于 l。如果否，则增加计数并打印该元素。 以下是上述办法的实施:
C++
// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to count the number of characters
// whose ascii value not in range [l, r]
int CountCharacters(string str, int l, int r)
{
// Initializing the count to 0
int cnt = 0;
// using map to print a character only once
unordered_map<char, int> m;
int len = str.length();
for (int i = 0; i < len; i++) {
// Increment the count
// if the value is less
if (!(l <= str[i] and str[i] <= r)) {
cnt++;
if (m[str[i]] != 1) {
cout << str[i] << " ";
m[str[i]]++;
}
}
}
// return the count
return cnt;
}
// Driver code
int main()
{
string str = "geeksforgeeks";
int l = 102, r = 111;
cout << "Characters with ASCII values"
" not in the range [l, r] \nin the given string are: ";
cout << "\nand their count is " << CountCharacters(str, l, r);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the above approach
import java.util.*;
class Solution
{
// Function to count the number of characters
// whose ascii value not in range [l, r]
static int CountCharacters(String str, int l, int r)
{
// Initializing the count to 0
int cnt = 0;
// using map to print a character only once
Map<Character, Integer> m= new HashMap<Character, Integer>();
int len = str.length();
for (int i = 0; i < len; i++) {
// Increment the count
// if the value is less
if (!(l <= str.charAt(i) && str.charAt(i) <= r)) {
cnt++;
if(m.get(str.charAt(i))!=null)
if (m.get(str.charAt(i)) != 1) {
System.out.print(str.charAt(i) + " ");
m.put(str.charAt(i),m.get(str.charAt(i))==null?0:m.get(str.charAt(i))+1);
}
}
}
// return the count
return cnt;
}
// Driver code
public static void main(String args[])
{
String str = "geeksforgeeks";
int l = 102, r = 111;
System.out.println( "Characters with ASCII values not in the range [l, r] \nin the given string are: ");
System.out.println( "\nand their count is " + CountCharacters(str, l, r));
}
}
//contributed by Arnab Kundu
Python 3
# Python3 implementation of the
# above approach
# Function to count the number of
# characters whose ascii value not
# in range [l, r]
def CountCharacters(str, l, r):
# Initializing the count to 0
cnt = 0
# using map to pra character
# only once
m = {}
length = len(str)
for i in range(0, length):
# Increment the count if the
# value is less
if (not(l <= ord(str[i]) and
ord(str[i]) <= r)):
cnt += 1
if ord(str[i]) not in m:
m[ord(str[i])] = 0
print(str[i], end = " ")
m[ord(str[i])] += 1
# return the count
return cnt
# Driver Code
if __name__ == '__main__':
str = "geeksforgeeks"
str = str.strip()
l = 102
r = 111
print("Characters with ASCII values", end = "")
print(" not in the range [l, r]\n",
"in the given string are: ", end = "")
print("\nand their count is ",
CountCharacters(str, l, r))
# This code is contributed by
# Shubham Singh(SHUBHAMSINGH10)
C
// C# implementation of the above approach
using System;
using System.Collections;
using System.Collections.Generic;
class GFG
{
// Function to count the number of characters
// whose ascii value not in range [l, r]
static int CountCharacters(string str, int l, int r)
{
// Initializing the count to 0
int cnt = 0;
// using map to print a character only once
Dictionary<char, int> m = new Dictionary<char, int>();
int len = str.Length;
for (int i = 0; i < len; i++)
{
// Increment the count
// if the value is less
if (!(l <= str[i] && str[i] <= r))
{
cnt++;
if(!m.ContainsKey(str[i]))
{
m[str[i]] = 0;
Console.Write(str[i] + " ");
}
m[str[i]]++;
}
}
// return the count
return cnt;
}
// Driver code
public static void Main(string []args)
{
string str = "geeksforgeeks";
int l = 102, r = 111;
Console.Write( "Characters with ASCII values" +
"not in the range [l, r] \nin" +
"the given string are: ");
Console.WriteLine( "\nand their count is " +
CountCharacters(str, l, r));
}
}
// This code is contributed by rutvik_56
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation of the above approach
// Function to count the number of characters
// whose ascii value not in range [l, r]
function CountCharacters($str, $l, $r)
{
// Initializing the count to 0
$cnt = 0;
// using map to print a character
// only once
$m = array_fill(0, 256, NULL);
$len = strlen($str);
for ($i = 0; $i < $len; $i++)
{
// Increment the count
// if the value is less
if (!($l <= ord($str[$i]) and
ord($str[$i]) <= $r))
{
$cnt++;
if (isset($m[ord($str[$i])]) != 1)
{
echo $str[$i] . " ";
$m[ord($str[$i])]++;
}
}
}
// return the count
return $cnt;
}
// Driver code
$str = "geeksforgeeks";
$l = 102;
$r = 111;
echo "Characters with ASCII values not in the " .
"range [l, r] \nin the given string are: ";
echo "\nand their count is " .
CountCharacters($str, $l, $r);
// This code is contributed
// by ChitraNayal
?>
java 描述语言
<script>
// JavaScript implementation of the above approach
// Function to count the number of characters
// whose ascii value not in range [l, r]
function CountCharacters(str,l,r)
{
// Initializing the count to 0
let cnt = 0;
// using map to print a character only once
let m= new Map();
let len = str.length;
for (let i = 0; i < len; i++) {
// Increment the count
// if the value is less
if (!(l <= str[i].charCodeAt(0) &&
str[i].charCodeAt(0) <= r)) {
cnt++;
if(!m.has(str[i]))
{
m.set(str[i],0);
document.write(str[i] + " ");
}
m.set(str[i],m.get(str[i]+1));
}
}
// return the count
return cnt;
}
// Driver code
let str = "geeksforgeeks";
let l = 102, r = 111;
document.write( "Characters with ASCII values " +
"not in the range [l, r] <br>in the given string are: ");
document.write( "<br>and their count is " +
CountCharacters(str, l, r));
// This code is contributed by avanitrachhadiya2155
</script>
Output:
Characters with ASCII values not in the range [l, r]
in the given string are: e s r
and their count is 7
时间复杂度: O(n)，其中 n 为字符串长度。
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