Question

可以说我有一个大小不一的字符串，是“ OOOO”。我需要以下代码还返回“ O0O0、0O0O，0OO0，O00O，O000，OO00，OOO0”。基本上，O的所有可能变化都替换为0。做到这一点的最佳方法是什么？当前仅返回“ 0OOO，00OO，000O，0000”

public class test {

    public static void main(String[] args) throws Exception {

        //Set the Desired Capabilities
        DesiredCapabilities caps = new DesiredCapabilities();
        caps.setCapability("deviceName", "Galaxy S9+");
        caps.setCapability("udid", "352402097079489"); //Give Device ID of your mobile phone
        caps.setCapability("platformName", "Android");
        caps.setCapability("platformVersion", "9.0");
        caps.setCapability("appPackage", "com.example.android.myAPP,com.android.settings");
        caps.setCapability("appActivity", "com.google.android.finsky.activities.MainActivity");
        caps.setCapability("noReset", "true");
        try {
            //  System.setProperty("webdriver.http.factory", "apache");
            AppiumDriver < MobileElement > driver = new AndroidDriver < MobileElement > (new URL("http://localhost:4723/wd/hub"), caps);

        } catch (MalformedURLException e) {
            System.out.println(e.getMessage());
        }
    }

}

Answer 1

一种方法是通过递归：

Public Iterator Function ParseOs(ByVal PlayerName As String) As IEnumerable(Of String)            
   Dim i as Integer = PlayerName.IndexOf("O")    
   If i = -1 Then
       Yield Return PlayerName
       Yield Break
   End If

   'Keep everything (if anything) *before* the O
   Dim base As String = PlayerName.SubString(0, i)

   Dim permutations As IEnumerable(Of String) = New String() {""}
   If PlayerName.Length > i Then
       'Recurse to find all the variations for everything *after* the O
       permutations = ParseOs(PlayerName.SubString(i+1))
   End If

   For Each permutation As String in permutations
        Yield Return base & "O" & permutation
        Yield Return base & "0" & permutation
   Next
End Function

如果您确定，输入将始终是“ O”，则可以进一步简化：

Public Iterator Function ParseOs(ByVal PlayerName As String) As IEnumerable(Of String)            
   Dim permutations As IEnumerable(Of String) = New String() {""}
   If PlayerName.Length > 1 Then
       permutations = ParseOs(PlayerName.SubString(1))
   End If

   For Each permutation As String in permutations
        Yield Return "O" & permutation
        Yield Return "0" & permutation
   Next
End Function

您可以通过将它们分成两个函数来提高效率，在这些函数中，递归部分处理字符数组而不是字符串。然后，通过将数组转换为使用Span<Char>，可以进一步提高的效率，因此不必为每个递归调用分配新的数组。

创建字符串排列以在O和0之间变化

1 个答案: