为什么我会超过时间限制？ 如何改善代码？

Question

目标是找到大于n的最小数字，并且该数字的数字总和可以完全除以11。在运行它之后，它什么也不返回。

public class M1
{
public static int nextCRC(int n)
{
    int crc;
    for (crc=n+1;crc!=0;crc++) 
    {
        int sum=0;
        for (crc =n+1 ; crc!=0; crc = crc / 10)
        {
            int digit = crc % 10;
            sum += digit;
            if ( sum %11 == 0)
            {
                break;
            }  
        }
    }       
    return crc;             
}
public static void main(String[] args)
{
    M1 Model=new M1();
    Model.nextCRC(20);
    System.out.println(Model.nextCRC(20));
}

}

Answer 1

干得好：

import java.math.*;

public class M1
{
public static int nextVal(int n)
{   
    int sum = 0;
    int temp = n;
    while(n>9){
        sum = sum + n%10;
        n = n/10;
    }
    sum = sum+n;
    if((sum%11) == 0){
        return temp;
    }
    return nextCRC(temp+1);
}

public static int nextCRC(int n) {
    return nextVal(n+1);
}

public static void main(String[] args)
{
    M1 Model=new M1();
    Model.nextCRC(20);
    System.out.println(Model.nextCRC(40));
}

}

Answer 2

这是我的解决方案：

import java.util.stream.LongStream;

/*
the goal is to find the least number that is greater than n and the digit sum of that number can be fully divided by 11.
 */
public class M1 {
  private static long nextCRC(int n) {
    return LongStream.iterate(n + 1, l -> ++l)
      .filter(l -> sumOfBase10Digits(l) % 11 == 0)
      .findFirst().getAsLong();
  }

  private static long sumOfBase10Digits(final long num) {
    if (num == 0) {
      return 0;
    } else {
      return num % 10 + sumOfBase10Digits(num / 10);
    }
  }

  public static void main(String... args) {
    System.out.println(nextCRC(20));
    System.out.println(nextCRC(200));
    System.out.println(nextCRC(98765));
    System.out.println(nextCRC(92));
  }
}