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为什么它在HashMap中返回null,而不是String? 爪哇

[英]Why it returns null, but not String in HashMap? java

How could I get values from Map?我如何从 Map 中获取值? I study Collections and wanna leave that order <String, Integer>.我研究 Collections 并想保留 <String, Integer> 的顺序。 I tried before <Integer, String> and in loop I put res.append(map.get(num));我之前尝试过 <Integer, String> 并在循环中放入res.append(map.get(num)); — that's help me. ——那是帮我。 But have no clue how to do it in reverse order.但是不知道如何以相反的顺序进行操作。 Also I wrote https://docs.oracle.com/javase/8/docs/api/java/util/HashMap.html#get-java.lang.Object- , but I know exactly that I have a key..我也写了https://docs.oracle.com/javase/8/docs/api/java/util/HashMap.html#get-java.lang.Object- ,但我确切地知道我有一个密钥..

import java.util.LinkedHashMap;
import java.util.Map;

public class Main {
    public static void main(String[] args) {
        int x = 555;
        String result = romanianConverter(x);
        System.out.println(result);
    }

    public static String romanianConverter(int value) {
        Map<String, Integer> map = new LinkedHashMap<>();
//        map.put(1000, "M");
//        map.put(900,"CM");
//        map.put(500,"D");
//        map.put(400,"CD");
//        map.put(100,"C" );
//        map.put(90,"XC" );
//        map.put(50,"L" );
//        map.put(40,"XL");
//        map.put(10,"X" );
//        map.put(9,"IX" );
//        map.put(5,"V" );
//        map.put(4,"IV" );
//        map.put(1,"I" );
        map.put("M", 1000);
        map.put("CM", 900);
        map.put("D", 500);
        map.put("CD", 400);
        map.put("C", 100);
        map.put("XC", 90);
        map.put("L", 50);
        map.put("XL", 40);
        map.put("X", 10);
        map.put("IX", 9);
        map.put("V", 5);
        map.put("IV", 4);
        map.put("I", 1);

        StringBuilder stringBuilder = new StringBuilder("Result = ");

        for (int number2 : map.values()) {
            while (number2 <= value) {
                stringBuilder.append(map.keySet().add(number2));
                value -= number2;
            }
        }
        return stringBuilder.toString();
    }
}

Traversing a HashMap is pretty straightforward.遍历HashMap非常简单。 First of all you have to understand that the first Class in the <> diamond brackets is the KEY and the second is the VALUE .首先,您必须了解<>菱形括号中的第一个 Class 是KEY ,第二个是VALUE

So if I create a new HashMap<Integer, String> the elements will be of the following form所以如果我创建一个新的HashMap<Integer, String>元素将是以下形式

1 -> "str",
2 -> "str2"
....

To traverse this HashMap And we want only the keys this is the way to do it:遍历这个HashMap并且我们只需要keys这是这样做的方式:

Map<String, Object> map = ...;

for (Integer key : map.keySet()) {
    // ...
}

If we want only the values:如果我们只想要值:

for (String value : map.values()) {
    // ...
}

If we want them both:如果我们同时想要它们:

for (Map.Entry<Integer, String> entry : map.entrySet()) {
    Integer key = entry.getKey();
    String value = entry.getValue();
    // ...
}

Still I am sure you are trying to conver integers in Roman Numerals.我仍然确定您正在尝试转换罗马数字中的integers This already has a solution here .在这里已经有了解决方案

In current implementation you need to iterate the map entries returned by Map::entrySet method to use the key as soon as the value is found:在当前实现中,您需要迭代Map::entrySet方法返回的映射条目,以便在找到值后立即使用该键:

// ...
    for (Map.Entry<String, Integer> me: map.entrySet()) {
        int number2 = me.getValue();
        while (number2 <= value) {
            stringBuilder.append(me.getKey());
            value -= number2;
        }
    }
// ...

Or, if the reverse map is implemented Map<Integer, String> map a keySet should be iterated:或者,如果实现了反向映​​射Map<Integer, String> map则应迭代keySet

        for (Integer number2: map.keySet()) {
            while (number2 <= value) {
                stringBuilder.append(map.get(number2));
                value -= number2;
            }
        }

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