TL; DR

Question

我的程序需要检查字符串中的3个连续字母（并检查整个字符串）。我可以让它以一种笨拙的方式检查它们，例如＆＃34;检查qwe＆＃34;，＆＃34;检查＆＃34; wer＆＃34;，检查＆＃34; ert＆＃34;，但这看起来很乱，做得很糟。

static void Main(string[] args)
{
    string BadLetters = "qwertyuiopasdfghjklzxcvbnm";
    string password = "Blablauio"; 

    for (int i = 1; i <= 30; i++)
    {
        // This checks if it contains "qwe" but i want it to 
        // cycle through the rest (such as "wer" or "rty")
        if (password.Contains(BadLetters.Substring(0, 3)))             {
            Console.WriteLine("password contains 3 consequtive letters in BadLetters");                    
        }
    }

    Console.ReadKey();
}

问题是这只会检查BadLetters（qwe）的前3个字母，而且它不会查找＆＃34; ert＆＃34;等。

Answer 1

如果循环密码变量会更好，如下所示：

    string badLetters = "qwertyuiopasdfghjklzxcvbnm";
    string password = "Blablauio"; 
    for (int i = 0; i < password.Length-2; i++)
    {
        if (badLetters.Contains(password.Substring(i,3))) 
        {
            Console.WriteLine("password contains 3 consequtive letters in BadLetters");
        }
    }

显然，您还必须检查密码是否至少为3个字符这个循环可能在键盘行交叉字母上失败，即＆＃34; opa＆＃34;或者＆＃34; pas＆＃34;，这应该被认为是正确的值，所以你可以这样做：

string badLettersR1 = "qwertyuiop";
string badLettersR2 = "asdfghjkl";
string badLettersR3 = "zxcvbnm";
string password = "Blablauio"; 
for (int i = 0; i < password.Length-2; i++)
{
    if (badLettersR1.Contains(password.Substring(i,3)) ||
        badLettersR2.Contains(password.Substring(i,3)) ||
        badLettersR3.Contains(password.Substring(i,3))) 
    {
        Console.WriteLine("password contains 3 consequtive letters in BadLetters");
    }
}

Answer 2

也许你可以尝试编辑你的字符串，所以3个字母都消失了，然后你就像以前一样做下一个字母。

或者您可以添加2个变量firstletter和secondletter 4，因此它会跳过前3个变量并重复以下变量

我知道你的字符串是连续3个......

Answer 3

您需要遍历password中的每个字符，在BadLetters中找到其索引，并检查password中的后两个字符是否与BadLetters中的后两个字符匹配。我还改变了for loop的停止条件，因为你只需要遍历password中的倒数第二个字符

string BadLetters = "qwertyuiopasdfghjklzxcvbnm";
string password = "Blablauio";
for (int i = 0; i < password.Length - 2; i++)
{
    var j = BadLetters.IndexOf(password[i]);
    if (j > -1 && j + 2 < BadLetters.Length &&
        password[i + 1] == BadLetters[j + 1] &&
        password[i + 2] == BadLetters[j + 2])
    {
        Console.WriteLine("password contains 3 consequtive letters in BadLetters");
    }
}

Answer 4

TL; DR

您可以浏览符号数组并比较它的索引，如下所示：

if (BadLetters.IndexOf(my_word[i]) - BadLetter.IndexOf(my_word[i-1]) == 1) {
     Console.WriteLine("Consequent letters detected!");
}

您可以计算后续字母，并在计数超过 3

时提醒

我提供了键盘中所有行的详细代码。你可以添加另一行（即大写）而无需修改任何代码。

您还可以控制N - 字符串中禁止的后续字符数。

还有Check方法仅用于演示工作结果：

q - ok
qw - ok
qwe - password contains 3 consequtive letters in BadLetters
abdfsk - ok
ehjk - password contains 3 consequtive letters in BadLetters
bnm - password contains 3 consequtive letters in BadLetters

代码

.net上的代码小提琴：https://dotnetfiddle.net/4oILkj

public static String[] KeyboardLines = new [] {
    "1234567890", 
    "qwertyuiop[]", 
    "asdfghjkl;'\\", 
    "`zxcvbnm,./"
};

public static Int32 GetLine(char c){
    for (int i = 0; i < KeyboardLines.Length; i++) {
        if (KeyboardLines[i].IndexOf(c) > -1) {
            return i;
        };
    }
    return -1;
}

public static bool HasConsequenceLetters(string str, int n = 3) {
    if (str.Length < n) { 
        return false;
    }

    char previousLetter = str[0];
    int previousLine = GetLine(previousLetter);
    int previousLetterIndex = KeyboardLines[previousLine].IndexOf(previousLetter);
    Int32 consequentLettersCount = 1;


    for (int i = 1; i < str.Length; i++) {
        var currentLetter = str[i];
        var currentLine = GetLine(currentLetter);
        var currentLetterIndex = KeyboardLines[currentLine].IndexOf(currentLetter);

        if (currentLine != -1 && currentLine == previousLine) {
            if (currentLetterIndex - previousLetterIndex == 1) {
                consequentLettersCount += 1;
            }
        }
        else {
            consequentLettersCount = 1;
        }

        if (consequentLettersCount == n) {
            return true;
        }

        previousLetter = currentLetter;
        previousLetterIndex = currentLetterIndex;
        previousLine = currentLine;
    }
    return false;
}

的改进

如果GetLine函数将返回带行号的字符索引，然后不仅比较字符而是比对（LetterLine，LetterIndex），则可以改进此方法。但这要求我们使用元组或类，但我不认为你真的想要这个。

Answer 5

我发现的一个简单的是

using System;

public class Program
{
    public static void Main()
    {
        string BadLetters = "qwertyuiopasdfghjklzxcvbnm";
        string password = "Blablauio";
        Console.WriteLine(password.IndexOf(BadLetters.Substring(0,3))>=0?"Present":"Not Present");
        Console.ReadLine();
    }
}

您可能需要检查两个字符串的空状态

如何检查字符串中的3个连续字母？

5 个答案:

TL; DR

代码

的改进