我已经编写了一个问题的代码，但它有很多if else循环。任何人都可以帮助我摆脱一些if else条件。谢谢 [英] i have written code for one problem but it have lots of if else loop. can anybody help me to get the rid of some if else condition. thank you

问题描述

问题是：输入3个数字作为输入（可以是1位或2位数字）。写一个程序，找到组合这3个数字可以形成的最小日期。这应该验证日历的所有逻辑约束（例如，我们不能在2月30日）





我的代码是：

problem is : Take 3 numbers as input (can be 1 or 2 digit numbers).Write a program to find the minimum date which can be formed by combining these 3 digits. This should validate for ALL logical constraints of a calendar (e.g. we cannot have 30th Feb)


my code is :

public class ThreeNum {
    public static void main(String[] args) {
        int d = 0, m = 0, y = 0;
        int num[] = new int[3];

        input = /* accept input from user */

        for(int i = 0; i < 3; i++) {
            System.out.print("Enter Number: ");
            num[i] = input.nextInt();
        }

  sort num  /* use of any sorting function to sorting array. */

        if (num[0] > 12) {
            System.out.print("Date is invalid. Try again");
            System.exit(0);
        } else if (num[2] <= 12) {
            y = num[0]; m = num[1]; d = num[2];
        } else if (num[2] <= 31) {
            if (num[1] <= 12) {
                y = num[0]; m = num[1]; d = num[2];
            } else {
                m = num[0]; y = num[1]; d = num[2];
            }
        } else {
            if (num[1] <= 31) {
                y = num[2]; d = num[1]; m = num[0];
            } else {
                System.out.print("Date is invalid. Try again");
                System.exit(0);
            }
        }

        if (m == 4 || m == 6 || m == 9 || m == 11) {
            if (d <= 30) {
                System.out.print("Corrent Date is " + d + "/" + m + "/" + y);
                System.exit(0);
            } else if (y <= 12) {
                if (y == 1 || y == 3 || y == 5 || y == 7 || y == 8 || y == 10 || y == 12)

{
                    m = m + y; y = m - y; m = m - y;
                    System.out.print("Corrent Date is " + d + "/" + m + "/" + y);
                    System.exit(0);
                }
            } else {
                d = d + y; y = d - y; d = d - y;
                System.out.print("Corrent Date is " + d + "/" + m + "/" + y);
                System.exit(0);
            }
        } else if (m == 2) {
            if (d <= 29) {
                if((y % 4) == 0) {
                    System.out.print("Corrent Date is " + d + "/" + m + "/" + y);
                    System.exit(0);
                } else if (y <= 28) {
                    d = d + y; y = d - y; d = d - y;
                    System.out.print("Corrent Date is " + d + "/" + m + "/" + y);
                    System.exit(0);
                } else {
                    System.out.print("Date is invalid. Try again");
                    System.exit(0);
                }
            } else if (d <= 31 & y <= 28) {
                d = d + y; y = d - y; d = d - y;
                System.out.print("Corrent Date is " + d + "/" + m + "/" + y);
                System.exit(0);
            } else {
                System.out.print("Date is invalid. Try again");
                System.exit(0);
            }
        } else {
            System.out.print("Corrent Date is " + d + "/" + m + "/" + y);
            System.exit(0);
        }
    }
}

推荐答案

这不仅仅是你的if..else丰富的问题 - 假设排序数字会给你正确的订单，你犯了一个根本性的错误。



例如，如果我下周给你的代码约会：4 2013年9月，您的系统会将其排序为

It's not just your if..else profusion that's the problem - you have made a fundamental mistake in assuming that sorting numbers will give you the right order.

For example, if I give your code a date next week: 4th Sept 2013, your system will sort it as

13
9
4

这将使您的首次测试失败，因为你假设最高的数字是月份 - 为什么？



我这样做的方式有点偷偷摸摸，可能不是你的导师想要的，但是...

我根本不会对数字进行排序。相反，我会将它们传递给一个方法，该方法生成独特的组合（只有六个，所以它很容易）作为一个字符串：

which will fail your first test, because you are assuming that the highest number is the month - why?

The way I would do it is a little sneaky, and probably not what your tutor wants, but...
I wouldn't sort the numbers at all. Instead, I would pass them to a method which generated the unique combinations (there are only six of them, so it's pretty easy) as a string:

static string[] GetCombination(int[] list)
    {
    string[] op = new string[6];
    int opIndex = 0;
    for (int i = 0; i < 3; i++)
        {
        for (int j = 0; j < 3; j++)
            {
            if (i != j)
                {
                for (int k = 0; k < 3; k++)
                    {
                    if (k != i && k != j)
                        {
                        op[opIndex++] = string.Format("{0:00}{1:00}{2:00}", list[i], list[j], list[k]);
                        }
                    }
                }
            }
        }
    return op;
    }

然后，我会使用一个循环使用DateTime.TryParseExact将每个转换为DateTime。一个简单的日期类型为您提供最低的有效日期，如果有的话。

Then, I would use a loop to convert each of these to a DateTime using DateTime.TryParseExact. A simple sort of the dates gives you the lowest valid date, if there are any.

这篇关于我已经编写了一个问题的代码，但它有很多if else循环。任何人都可以帮助我摆脱一些if else条件。谢谢的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！