通过递归仅获取数字中的奇数位

如何解决通过递归仅获取数字中的奇数位

所以我的问题是，我有一个像123这样的数字，并且因为pelip表示我希望结果为13。

问题是，首先使用im的方法im将获得求反结果（例如31），其次im在不应该存在的末尾得到零，而不是将数字相加并将其求和我不明白为什么。

所以要澄清一下：

我的输出：

def apenas_digitos_impares(n):
    if n == 0:
        return 0
    elif (n%10)%2 == 0:
        return apenas_digitos_impares(n//10)
    elif (n%10)%2 == 1:
        return 10*(n%10) + apenas_digitos_impares(n//10)

正确的输出：

>>> apenas_digitos_impares(123)
40

程序：

>>> apenas_digitos_impares(123)
13

解决方法

您可以按照-

def apenas_digitos_impares(n):
    if n == 0:
        return 0
    elif (n%10)%2 == 0:
        return apenas_digitos_impares(n//10)
    elif (n%10)%2 == 1:
        return (n%10) + 10*apenas_digitos_impares(n//10)
        
print(apenas_digitos_impares(123))

输出：

13

,

• 解决方案＃1： 您可以使用一种方法进行操作

---

public class MyScheduler {

    @Autowired
    private JobLauncher customerJobLauncher;
    
    @Autowired
    private JobLauncher abcJobLauncher;
    
    @Autowired
    private JobLauncher xyzJobLauncher;

    @Autowired
    @Qualifier(value = "customerJob")
    private Job customerJob;
    
    @Autowired
    @Qualifier(value = "abcJob")
    private Job abcJob;
    
    @Autowired
    @Qualifier(value = "xyzJob")
    private Job xyzJob;

    @Scheduled(cron = "0 0 */1 * * *") // run at every hour for testing 
    public void handle() {
        JobParameters params = new JobParametersBuilder()
                .addString("cust.job.id",String.valueOf(System.currentTimeMillis()))
                .addDate("cust.job.date",new Date()).toJobParameters();
        long diff = 0;
        try {
            JobExecution jobExecution = customerJobLauncher.run(customerJob,params);
            Date start = jobExecution.getCreateTime();

            
            JobParameters job2Params = new JobParametersBuilder()
                    .addString("abc.job.id",String.valueOf(System.currentTimeMillis()))
                    .addDate("abc.job.date",new Date()).toJobParameters();
            
            JobExecution job2Execution = abcJobLauncher.run(abcJob,job2Params);            
            
            JobParameters job3Params = new JobParametersBuilder()
                    .addString("xyz.job.id",String.valueOf(System.currentTimeMillis()))
                    .addDate("xyx.job.date",new Date()).toJobParameters();
            
            JobExecution job3Execution = xyzJobLauncher.run(xyzJob,job3Params);
            
            Date end = job3Execution.getEndTime();
            diff = end.getTime() - start.getTime();

            log.info(JobExecutionTimeCalculate.getJobExecutionTime(diff));
        } catch (JobExecutionAlreadyRunningException | JobRestartException | JobInstanceAlreadyCompleteException
                | JobParametersInvalidException e) {
            log.error("Job Failed : " + e.getMessage());
        }
    }
}

输出：

import math


def convert_int_to_list(n):
    result = []

    while n > 0:
        if (n % 10) % 2 == 1:
            result.append(n % 10)
        n = math.floor(n / 10)

    return result


print(convert_int_to_list(1345986))

• 解决方案＃2 您不必更改方法，只需在方法之前声明一个数组

---

[9,5,3,1]

• 您需要奇数，然后将数组放置在正确的位置。

---

result = []

• 打印资源

---

 elif (n%10)%2 == 1:
      res.append(n%10)
      return 10*(n%10) + apenas_digitos_impares(n//10)

• 对于各种示例：

---

[3,1]

apenas_digitos_impares(1235)
apenas_digitos_impares(12356789)