字典顺序-两个char数组作为参数，寻找更好的解决方案

Question

我必须编写一个需要 2 char[]s 和 returns:

的函数

-1 如果第一个单词按字典顺序排在第二个单词之前
0 如果它们是同一个词
1如果在

我知道 compareTo() 方法，但这是一项作业，我需要避免使用它。到目前为止，我的代码运行良好，我用不同的词做了一些测试。

我想知道是否有另一种方法可以做到这一点，我的代码感觉没有优化，它又长又重复：

public static int lexico(char[] word1, char[] word2) {
        int length1 = word1.length;
        int length2 = word2.length;

        if (length1 == length2) {
            for (int i = 0; i < length1; i++) {
                if (word1[i] < word2[i]) {
                    return -1;
                } else if (word1[i] > word2[i]) {
                    return 1;
                } else if (i == length1 - 1) {
                    return 0;
                }
            }
        }

        if (length1 < length2) {
            for (int i = 0; i < length1; i++) {
                if (word1[i] < word2[i]) {
                    return -1;
                } else if (word1[i] > word2[i]) {
                    return 1;
                } else if (i == length1 - 1) {
                    // If I'm here then it means that all of the characters
                    // from 0 to length1-1 are equals
                    // but since length of the first string is shorter than the second,
                    // the first string will be put before the second
                    return -1;
                }

            }
        }

        if (length1 > length2) {
            for (int i = 0; i < length2; i++) {
                if (word1[i] < word2[i]) {
                    return -1;
                } else if (word1[i] > word2[i]) {
                    return 1;
                } else if (i == length1 - 1) {
                    return 1;
                }
            }
        }

        return -999;
    }

public static void main(String[] args) {
    char[] share = { 's', 'h', 'a', 'r', 'e' };
    char[] ship = { 's', 'h', 'i', 'p' };

    System.out.println(lexico(share, ship)); // -1 share is before ship
    System.out.println(lexico(ship, share)); // 1 ship is after share
    System.out.println(lexico(ship, ship)); // 0 same word

}

Answer 1

给您的几点说明：

你只需要一个循环：从开始到两个长度中较低的一个。如果数组相同直到两个长度中的较小者并且它们的长度不同，你的分配应该告诉你什么 return （通常它会是 -1 如果左边的数组比右边的短，1否则）。
a < b 不是两个字符的有效字母比较（大多数程序员说 "lexicographic" 时的意思，"lexico" 意思是 "pertaining to words"), 这是一个 numeric 比较。现在，String 的 compareTo claims to use "lexicographic ordering," but it really just uses numeric ordering, so that may be good enough for what you're doing. If you want alphabetic ordering, I can't think of a JDK comparison method that accepts two single chars to compare rather than strings. There may be one I don't know of, or you may have to create one-character strings to do the comparison (with a Collator) 将（例如）正确识别 "voilà" 中的 à 应该在其中任何其他字母之前。

Answer 2

好吧，我放弃并写了它...没有单元测试（总是交付代码单元测试）但我认为这应该给你一个很好的功能工作想法（有像不同长度等边缘情况等。我还没有处理过，但你的想法是对的吗？）...干杯并玩得开心:=）...我想使用流可以更有效地完成...

public static void main(String[] args) {
    List<Character> firstCharArr = new ArrayList<>(Arrays.asList('a', 'c'));
    List<Character> secondCharArr = new ArrayList<>(Arrays.asList('x', 'c'));

    BiFunction<Character, Character, Integer> charCompareFunction = new BiFunction<Character, Character, Integer>() {
        @Override
        public Integer apply(Character character, Character character2) {
            if (character.charValue() < character2.charValue()) {
                return -1;
            } else if (character.charValue() == character2.charValue()) {
                return 0;
            } else {
                return 1;
            }
        }
    };

    int i = 0;
    for (Character firstLineChar : firstCharArr) {
        if (charCompareFunction.apply(firstLineChar, secondCharArr.get(i++)) == 0) {
            continue;
        } else {
            if (charCompareFunction.apply(firstLineChar, secondCharArr.get(i++)) < 0) {
                System.out.println("FirstLine is Smaller");
                break;
            } else {
                System.out.println("FirstLine is Larger");
                break;
            }
        }
    }
}

其实可以利用上面的方法，证明Java的字符字典序：字母前的数字大写先于小写

字典顺序-两个char数组作为参数，寻找更好的解决方案

Lexicographical order - two char array as parameters, looking for a better solution

java

lexicographic