由于数据类型的原因，循环在计算之前退出

Question

我的程序规范如下。1.所有四位数都不同2.千位中的数字是十位中数字的三倍3.数字是奇数。4.数字之和是27。我遗漏了整个程序的一些代码。它有一个干净的编译，但当它运行时会自动终止。我认为问题出在数据类型的转换上。

int randomNumber = rand.nextInt(9000) + 1000;
String randomString;
boolean found = true;   

while (found)
{   

// converting to string to find position of digits and value        
randomString = String.valueOf(randomNumber);

// converting to char to transfer back to in while knowing the position of the digits 
char position0a = randomString.charAt(0);
char position1a = randomString.charAt(1);
char position2a = randomString.charAt(2);
char position3a = randomString.charAt(3);

// coverted back to int
int position0 = Character.getNumericValue(position0a);
int position1 = Character.getNumericValue(position1a);
int position2 = Character.getNumericValue(position2a);
int position3 = Character.getNumericValue(position3a);

int sumCheck = position0a + position1a + position2a + position3a;
int digit30Check = 3 * position2;

//checking addition to 27
String sumCheck27 = "27";
String sumCheck28 = String.valueOf(sumCheck);

// checking all digits are different
if (position0 != position1 && position0 != position2 && position0 != position3  &&

position1 != position2 && position1 != position3 && position2 != position3) 
{
if (position3 != digit30Check) // check if the digit in the thousands place 3 * tens
{
    if (sumCheck27.equals(sumCheck28)) // check if the sum is 27
    {
        if (position0 != 1 && position0 != 3 
&& position0 != 5 && position0 != 7 &&      
position0 != 9 && position1 != 1 && position1 != 3 
&& position1 != 5 && position1 != 7 && 
position1 != 9 && position2 != 2 && position2 != 3 
&& position2 != 5 && position2 != 7 && 
position2 != 9 && position3 != 3 && position3 != 3 && 
position3 != 5 && position3 != 7 && position3 != 9)
        { 
// checks for odd digits
         found = false;
         System.out.println(randomNumber);

        }
        else 
        randomNumber = rand.nextInt(9000) + 1000;
    }
    else 
    randomNumber = rand.nextInt(9000) + 1000;               
}
else 
randomNumber = rand.nextInt(9000) + 1000; 
}
else 
 randomNumber = rand.nextInt(9000) + 1000; 

 // end while
 }

Answer 1

我相信这可以通过观察给定的不等式来大大简化，比如

// thousands + hundreds + tens + ones == 27
// thousands = tens * 3, or tens = thousands / 3
// thousands = 3, 6 or 9
public static void main(String[] args) {
    // thousands = tens * 3; so it must be a multiple of 3.
    for (int thousands = 3; thousands < 10; thousands += 3) {
        for (int hundreds = 0; hundreds < 10; hundreds++) {
            if (hundreds == thousands) {
                continue;
            }
            // thousands = tens * 3; so the reverse is also true
            int tens = thousands / 3;
            if (tens == hundreds) {
                continue;
            }
            // All of the digits sum to 27.
            int ones = 27 - thousands - hundreds - tens;
            if (ones > 9 || ones % 2 != 0 || ones == tens
                    || ones == hundreds || ones == thousands) {
                continue;
            }
            int val = (thousands * 1000) + (hundreds * 100) + (tens * 10)
                    + ones;
            System.out.println(val);
        }
    }
}

当然，输出仍然是

9738

Answer 2

将while（找到）交换为while（！找到）。

逻辑地思考一下。你不想在找到东西的时候去看。你想在找不到的时候继续找。

Answer 3

boolean found = false;  

while (found)

仅此一项就确保了while循环永远不会被输入，因为find是false。while循环中的任何内容都没有任何区别，因为它永远不会被执行。

你可能是想写

while (!found)

除了这个错误，你的情况也太复杂了。以下是简化它们的方法：

if ((position0 == (3 * position2)) && // note that position0 is the "thousands place", not position3
    ((position0+position1+position2+position3) == 27) && // sum of digits
    (position3 % 2 == 1) && // odd number
    (position0 != position1 && position0 != position2 && position0 != position3  &&
     position1 != position2 && position1 != position3 && position2 != position3)) { // different digits
    found = true;
}