检查给定的字符串是否为异码
原文:https://www . geesforgeks . org/check-when-given-string-heterogram-not/
给定一根绳子 S 。任务是检查给定的字符串是否是杂凑图。一个异形词是一个单词、短语或句子，其中字母表中没有一个字母出现超过一次。
例:
Input : S = "the big dwarf only jumps"
Output : Yes
Each alphabet in the string S is occurred
only once.
Input : S = "geeksforgeeks"
Output : No
Since alphabet 'g', 'e', 'k', 's' occurred
more than once.
想法是创建一个大小为 26 的散列数组，初始化为 0。遍历给定字符串的每个字母表，并在相应的哈希数组位置标记 1。如果第一次遇到该字母表，则返回 false。
下面是该方法的实现:
C++
// C++ Program to check whether the given string is Heterogram or not.
#include<bits/stdc++.h>
using namespace std;
bool isHeterogram(char s[], int n)
{
int hash[26] = { 0 };
// traversing the string.
for (int i = 0; i < n; i++)
{
// ignore the space
if (s[i] != ' ')
{
// if already encountered
if (hash[s[i] - 'a'] == 0)
hash[s[i] - 'a'] = 1;
// else return false.
else
return false;
}
}
return true;
}
// Driven Program
int main()
{
char s[] = "the big dwarf only jumps";
int n = strlen(s);
(isHeterogram(s, n))?(cout << "YES"):(cout << "NO");
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java Program to check whether the
// given string is Heterogram or not.
class GFG {
static boolean isHeterogram(String s, int n)
{
int hash[] = new int[26];
// traversing the string.
for (int i = 0; i < n; i++)
{
// ignore the space
if (s.charAt(i) != ' ')
{
// if already encountered
if (hash[s.charAt(i) - 'a'] == 0)
hash[s.charAt(i) - 'a'] = 1;
// else return false.
else
return false;
}
}
return true;
}
// Driver code
public static void main (String[] args)
{
String s = "the big dwarf only jumps";
int n = s.length();
if(isHeterogram(s, n))
System.out.print("YES");
else
System.out.print("NO");
}
}
// This code is contributed by Anant Agarwal.
Python 3
# Python3 code to check
# whether the given
# string is Heterogram
# or not.
def isHeterogram(s, n):
hash = [0] * 26
# traversing the
# string.
for i in range(n):
# ignore the space
if s[i] != ' ':
# if already
# encountered
if hash[ord(s[i]) - ord('a')] == 0:
hash[ord(s[i]) - ord('a')] = 1
# else return false.
else:
return False
return True
# Driven Code
s = "the big dwarf only jumps"
n = len(s)
print("YES" if isHeterogram(s, n) else "NO")
# This code is contributed by "Sharad_Bhardwaj".
C
// C# Program to check whether the
// given string is Heterogram or not.
using System;
class GFG {
static bool isHeterogram(string s, int n)
{
int []hash = new int[26];
// traversing the string.
for (int i = 0; i < n; i++)
{
// ignore the space
if (s[i] != ' ')
{
// if already encountered
if (hash[s[i] - 'a'] == 0)
hash[s[i] - 'a'] = 1;
// else return false.
else
return false;
}
}
return true;
}
// Driver code
public static void Main ()
{
string s = "the big dwarf only jumps";
int n = s.Length;
if(isHeterogram(s, n))
Console.WriteLine("YES");
else
Console.WriteLine("NO");
}
}
// This code is contributed by Vt_m.
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP Program to check
// whether the given string
// is Heterogram or not.
function isHeterogram($s, $n)
{
$hash = array();
for($i = 0; $i < 26; $i++)
$hash[$i] = 0;
// traversing the string.
for ($i = 0; $i < $n; $i++)
{
// ignore the space
if ($s[$i] != ' ')
{
// if already encountered
if ($hash[ord($s[$i]) -
ord('a')] == 0)
$hash[ord($s[$i]) -
ord('a')] = 1;
// else return false.
else
return false;
}
}
return true;
}
// Driven Code
$s = "the big dwarf only jumps";
$n = strlen($s);
if (isHeterogram($s, $n))
echo ("YES");
else
echo ("NO");
// This code is contributed by
// Manish Shaw(manishshaw1)
?>
java 描述语言
<script>
// Javascript program to check whether
// the given string is Heterogram or not.
function isHeterogram(s, n)
{
var hash = Array(26).fill(0);
// Traversing the string.
for(var i = 0; i < n; i++)
{
// Ignore the space
if (s[i] != ' ')
{
// If already encountered
if (hash[s[i].charCodeAt(0) -
'a'.charCodeAt(0)] == 0)
hash[s[i].charCodeAt(0) -
'a'.charCodeAt(0)] = 1;
// Else return false.
else
return false;
}
}
return true;
}
// Driver code
var s = "the big dwarf only jumps";
var n = s.length;
(isHeterogram(s, n)) ? (document.write("YES")) :
(document.write("NO"));
// This code is contributed by rutvik_56
</script>
输出:
YES
时间复杂度: O(N)
辅助空间: O(26)
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