我被困在这里.我只是继续制作新的字符串并将其转换为int还是我们有更快更好的方法?
I’m stuck here. Do I just keep making new strings and turn them to int or us there a faster better way?
public void biggest(int a){ int random; String aS = String.valueOf(a); int ah=9; if (a<10) System.out.println(a); for(int i= 0;i<aS.length();i++){ String firstNum = aS.substring(i,i+1); for (int j = ah; j > Integer.parseInt(firstNum); j--){ System.out.println(ah); } } } ``` 推荐答案在这种情况下，无需使用转换为String的方法，您可以通过对输入的余数进行模10取余，然后将输入除以来从输入数字中获取数字.数字乘以10并在数字>0.
There's no need to use conversion to String in this case, you can get the digits from the input number by getting a remainder by modulo 10, then dividing the input number by 10 and repeat it while the number > 0.
每个数字应存储在数组或列表中.
Each digit should be stored in an array or list.
要获取最多的这些数字，您应该对它们进行排序(例如 Arrays.sort 或 Collections.sort 这样的标准工具就可以了)，然后单击重新组装"将最低位数乘以1、10、100等，然后将其相加即可得出最大的数字.
To get the biggest number of these digits you should just sort them (standard facilities such as Arrays.sort or Collections.sort will do fine) and then "re-assemble" the biggest number from the lowest digit by multiplying it by 1, 10, 100, etc. and summing up.
因此，简单的实现可能如下:
So, plain implementation could be as follows:
public static int biggestPlain(int a) { List<Integer> digits = new ArrayList<>(); while (a > 0) { digits.add(a % 10); a /= 10; } Collections.sort(digits); int p = 1; int num = 0; for (int digit : digits) { num += p * digit; p *= 10; } return num; }此外，可以使用Stream API和lambda并应用相同的方法来实现此任务:
Also, this task can be implemented using Stream API and lambda and applying the same approach:
public static int biggestStream(int a) { AtomicInteger p = new AtomicInteger(1); // accumulate powers of 10 return IntStream.iterate(a, n -> n > 0, n -> n / 10) // divide input number by 10 while it > 0 .map(i -> (i % 10)) // get the digit .sorted() // sort (the lower digits first) .map(i -> p.getAndUpdate((x) -> x * 10) * i) // same as p * digit above .sum(); // get the result number }更新 在从"9"到"0"的数字之间进行迭代，并检查输入数字的字符串表示中是否可用这些数字.
Update Iterate over digits from '9' till '0' and check if they are available in the string presentation of the input number.
String 的解决方案:
public static void biggest(int a) { String aS = String.valueOf(a); if (a < 10) { System.out.println(a); } String num = ""; int count = 0; out: for (char i = '9'; i >= '0'; i--) { for (int j = 0; j < aS.length(); j++) { char digit = aS.charAt(j); if (digit == i) { num += digit; if (++count == aS.length()) { break out; } } } } System.out.println(num + " / " + Integer.parseInt(num)); }