java - 如何根据单词范围检查输入单词的位置？

Question

如何检查输入的单词是否位于给定“单词范围”的“之前”、“内部”或“之后”？对于每个输入的单词，我们只需要输出 "before", "inside" or "after" ..

判断顺序，字符优先规则为:'' < 'A' < 'a' < 'B' < 'b' .. 'Z' < 'z' .

Format of input is:
1) <start word> <end word>
2) A sequence of N words

例如:

Input:
============
Apple Pear
Aa
Aq
App
Apple

Output:
============
before // as Aa < Apple
inside // as Aq > Apple && Aq < Pear
before // as App < Apple
inside // as Apple == Apple && Apple < Pear

下面是我得到的，但它给出了错误的输出，所以我的比较逻辑看起来是错误的。

public static void main(String args[] ) throws Exception {
    Scanner sc = new Scanner(System.in);
    String startWord =  sc.next(); // this will give "Apple"
    String endWord =  sc.next(); // this will give "Pear"
    while (sc.hasNext()) {
       // below will give "Aa", or "Aq", or "App" or "Apple" one by one
        String queryWord =  sc.next();
        if(queryWord.compareTo(startWord) < 0 && queryWord.compareTo(endWord) < 0) {
            System.out.println("before");
        } else if(queryWord.compareTo(startWord) > 0 && queryWord.compareTo(endWord) > 0) {
            System.out.println("after");
        } else if(queryWord.compareTo(startWord) == 0 && queryWord.compareTo(endWord) == 0) {
            System.out.println("inside");
        }
    }
    sc.close();
}

Update:

那么逻辑会是这样吗？

    while (sc.hasNext()) {
        String queryWord =  sc.next();
        int qValue = 0;
        for(char c : queryWord.toCharArray()) {
            qValue += map.get(c);
        }
        if(qValue > sValue && qValue < eValue) {
            System.out.println("inside");
        } else if(qValue > sValue && qValue > eValue) {
            System.out.println("before");
        } else {
            System.out.println("after");
        }
    }

Answer 1

在发布解决方案之前，让我们看一下输入范围。

单词的每个字母都是 [A,Z] 或 [a-z] 内的任何字母。

下面是我想用来计算给定输入是-before、inside 还是outside 的逻辑.

1. Let's initialize two variables - startWord and endWord.
2. Take input for the above two variables.
3. Now, I create a map which stores value for each letter constant. As per the question, my values will look something like this:-
   A - 1
   a - 2
   B - 3
   b - 4
   .
   .
   Z = 51
   z = 52.
4. Now I will calculate value of each word, so for example, value of Apple will be = A+p+p+l+e = 1+32+32+24+10 = 99.
5. Similarly I will do for Pear.
6. Now for each user input value, for example Aa. I will make this calculation again, so for Aa its = A+a = 1+2 = 3. And 3 is less then 99. 
7. Now it all comes to Maths, if this value is within the range of startWord and endWord, then I print inside, if its less then answer is before otherwise after.

希望这对您有所帮助!