作者：三星anycall | 来源：互联网 | 2023-09-24 18:03
检查所有给定的字符串是否都是等值线
原文:https://www . geeksforgeeks . org/check-如果所有给定的字符串都是-isogram-or-not/
给定一个包含 N 字符串的数组 arr ，任务是检查所有字符串是否都是等值线图。如果是，打印是，否则打印否。
一个等值线是一个字母出现不超过一次的单词。
示例:
输入:arr[]= {“ABCD”、“derg”、“erty”}
T3】输出:是
输入:arr[]= {“agka”、“lkmn”}
T3】输出:否
方法:如果字符串中没有字母出现超过一次，则该字符串为等值线图。现在为了解决这个问题，
- 遍历数组 arr ，对于每个字符串
- 创建字符的频率图。
- 任何字符频率大于 1 的地方，打印否并返回。
- 否则，遍历整个数组后，打印是。
下面是上述方法的实现:
C++
// C++ code for the above approach
#include
using namespace std;
// Function to check if a string
// is an isogram or not
bool isIsogram(string s)
{
vector<int> freq(26, 0);
for (char c : s) {
freq++;
if (freq > 1) {
return false;
}
}
return true;
}
// Function to check if array arr contains
// all isograms or not
bool allIsograms(vector<string>& arr)
{
for (string x : arr) {
if (!isIsogram(x)) {
return false;
}
}
return true;
}
// Driver Code
int main()
{
vector<string> arr = { "abcd", "derg", "erty" };
if (allIsograms(arr)) {
cout << "Yes";
return 0;
}
cout << "No";
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program for the above approach
import java.io.*;
class GFG {
// Function to check if a string
// is an isogram or not
static boolean isIsogram(String s)
{
int freq[] = new int[26];
char S[] = s.toCharArray();
for (char c : S) {
freq++;
if (freq > 1) {
return false;
}
}
return true;
}
// Function to check if array arr contains
// all isograms or not
static boolean allIsograms(String arr[])
{
for (String x : arr) {
if (isIsogram(x) == false) {
return false;
}
}
return true;
}
// Driver Code
public static void main(String[] args)
{
String arr[] = { "abcd", "derg", "erty" };
if (allIsograms(arr) == true) {
System.out.println("Yes");
}
else {
System.out.println("No");
}
}
}
// This code is contributed by Potta Lokesh
Python 3
# Python code for the above approach
# Function to check if a string
# is an isogram or not
def isIsogram (s):
freq = [0] * 26
for c in s:
freq[ord(c) - ord('a')] += 1
if (freq[ord(c) - ord('a')] > 1):
return False
return True
# Function to check if array arr contains
# all isograms or not
def allIsograms (arr):
for x in arr:
if (not isIsogram(x)):
return False
return True
# Driver Code
arr = ["abcd", "derg", "erty"]
if (allIsograms(arr)):
print("Yes")
else:
print("No")
# This code is contributed by Saurabh Jaiswal
C
// C# program for the above approach
using System;
public class GFG {
// Function to check if a string
// is an isogram or not
static bool isIsogram(String s)
{
int []freq = new int[26];
char []S = s.ToCharArray();
foreach (char c in S) {
freq++;
if (freq > 1) {
return false;
}
}
return true;
}
// Function to check if array arr contains
// all isograms or not
static bool allIsograms(String []arr)
{
foreach (String x in arr) {
if (isIsogram(x) == false) {
return false;
}
}
return true;
}
// Driver Code
public static void Main(String[] args)
{
String []arr = { "abcd", "derg", "erty" };
if (allIsograms(arr) == true) {
Console.WriteLine("Yes");
}
else {
Console.WriteLine("No");
}
}
}
// This code is contributed by 29AjayKumar
java 描述语言
<script>
// Javascript code for the above approach
// Function to check if a string
// is an isogram or not
const isIsogram = (s) => {
let freq = new Array(26).fill(0);
for (let c in s) {
freq[s.charCodeAt(c) - 'a'.charCodeAt(0)]++;
if (freq[s.charCodeAt(c) - 'a'.charCodeAt(0)] > 1) {
return false;
}
}
return true;
}
// Function to check if array arr contains
// all isograms or not
const allIsograms = (arr) => {
for (let x in arr) {
if (!isIsogram(arr[x])) {
return false;
}
}
return true;
}
// Driver Code
let arr = ["abcd", "derg", "erty"];
if (allIsograms(arr))
document.write("Yes");
else
document.write("No");
// This code is contributed by rakeshsahni
script>
Output
时间复杂度: O(N*M)，其中 N 是数组的大小，M 是最长字符串的大小
T3】辅助空间: O(1)
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