再試一次 Do-While 循環

Question

這是一個非常簡單的程序。 計算機和用戶在 0-3 之間選擇一個數字。 如果用戶沒有猜出與計算機相同的數字，我希望程序循環返回。

    String input; // input is a string variable
    int cpucrazy;
    int convertstring; 

// 第一步 //

    input = JOptionPane.showInputDialog("Guess a number between 0-3");
    convertstring = Integer.parseInt(input);

// 隨機部分 //

    Random ran = new Random ();
    cpucrazy = ran.nextInt(3);

// 計算！

if (cpucrazy < convertstring) {
     JOptionPane.showInputDialog(null, "Your guess is too high. Guess again?"); }


else if (cpucrazy > convertstring) {
     JOptionPane.showInputDialog(null, "Your guess is too low. Guess again?"); }


else JOptionPane.showMessageDialog(null, "Correct!!"); 

Answer 1

if (cpucrazy < convertstring) {
 JOptionPane.showInputDialog(null, "Your guess is too high. Guess again?"); }


else if (cpucrazy > convertstring) {
     JOptionPane.showInputDialog(null, "Your guess is too low. Guess again?"); }


else JOptionPane.showMessageDialog(null, "Correct!!"); 

您需要在上面的代碼周圍放置一個 while 循環（或某種類型的循環構造）並在以下情況下退出構造： cpucrazy == convertstring或在cpucrazy != convertstring停留在循環構造中

循環可能類似於以下偽代碼：

b = random();

do
{
     input a;
     if( a != b )
     {
         if(a < b )
         {
              print "Too low";
         }
         else
         {
              print "Too high";
         }
     }
} while( a != b );

print "You got it correct"