需要帮助最小和最大功能？

我有一个应用程序的数字猜测游戏，用户必须猜测一个0到100之间的数字，当他们猜出正确的程序询问他们是否想在他们完成游戏时再次玩游戏时，我会显示最少数量的猜测游戏和猜谜游戏中的最大数量。现在我所得到的是在使用“Math.min（，）”时所有猜测的总和？如何获得最小功能？功能代码进一步在下面。需要帮助最小和最大功能？

leastNumGuesses = Math.min（leastNumGuesses，猜测）;

double rightNum = Math.random() *100; 
int randomNum = (int) rightNum; //convert the random number to int 
int tries = 0; //single game gussess output 
int numberOfGames = 0; 
int allTries = 0; //accumalates all tries(sum of all tries) 
int guesses = 0; // guesses of all games combined 
int gameGuesses = 0; 
int leastNumGuesses = 100; 
int mostNumGuesses = 0; 


    while (choice.equalsIgnoreCase("y"))

{

System.out.println(); 

    int guess = getIntWithinRange(sc,"Enter the Number: ", 0, 100); 
    tries++; 
    guesses++; 
    gameGuesses++; 

    if (guess == randomNum) 
    { 
     numberOfGames++; 
     System.out.println("You got it in " + tries + " tries."); 
     leastNumGuesses = Math.min(leastNumGuesses,gameGuesses); 
     if (tries <=3) 
       System.out.println("Great work! You are a mathematical wizard."); 
     else if (tries > 3 && tries <= 7) 
       System.out.println("Not too bad! You've got some potential."); 
     else if (tries > 7) 
       System.out.println("What took you so long? Maybe you should take some lessons."); 

     System.out.println();  
     System.out.println("Would you like to play again (y/n):"); 
     choice = sc.nextLine(); 

    while (!choice.equalsIgnoreCase("n") && !choice.equalsIgnoreCase("y")) 
    { 

    System.out.println("Error! entry must be \"y\" or \"n\"."); 
    System.out.println("Would you like to play again (y/n):"); 
    choice = sc.nextLine(); 
    } 

     if (choice.equalsIgnoreCase("y"))  
     { // reset the random number & tries 

      rightNum = Math.random() *100; 
      randomNum = (int) rightNum; 
      tries=0; 
      gameGuesses++; 
     } 
     else if (choice.equalsIgnoreCase("n")) 
     {  
      allTries += guesses; 
      int averageNumGuess = allTries/numberOfGames; 

      System.out.println("Bye - Come back again"); 
      System.out.println("Number of Games Played: " + numberOfGames); 
      System.out.println("Average Number of Guesses: " + averageNumGuess); 
      System.out.println("Least Amount of Guesses In a Single Game: " + leastNumGuesses); 

     } 

    }

来源

2013-03-05 babaysteps

你能发布一个最小的代码片段：1）演示了这个问题，2）是可编译的？ – apnorton 2013-03-05 02:32:13

编译和运行代码时会发生什么？它与你想要和期望的有什么不同？ – 2013-03-05 02:50:36

@Code-Guru它为每个游戏添加所有猜测并将其显示为单个游戏中最少的猜测次数;我希望它能够在每场比赛中进行所有猜测，并在比赛中找到最小或最少的猜测。 – babaysteps 2013-03-05 02:54:37

看来你改变你想guesses在节目中代表什么。

记住guesses是总数的猜测在所有场比赛，而leastNumGuesses最初设置为100。在大多数情况下，您会发现guesses < leastNumGuesses，因此Math.min(guesses, leastNumGuesses)函数将返回guesses。

修复：使用除guesses之外的变量，例如gameGuesses来跟踪游戏中进行了多少次猜测。然后，Math.min(,)将按照您的预期行事。

来源

2013-03-05 02:37:50 apnorton

我试过了但是问题是min函数加上所有的猜测，也许是把变量放在错误的地方？我不知道，但这真的让我感到困扰。 – babaysteps 2013-03-05 02:49:26

@ user1852564在循环内放置'Math.min（guesses，leastNumGuesses）'行，以便每次游戏结束时执行，无论用户是否回答“y”或“n”。 – apnorton 2013-03-05 02:57:43

需要帮助最小和最大功能？

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