字符串基数排序 - StringIndexOutOfBoundsEception

Question

我正在编写自己的基数排序方法来对字符串中的单词进行排序（the big black cat sat on the  beautiful brown mat 将被排序为beautiful big black brown cat mat on sat the the）。该方法接收单个单词的 List（我自己的 List 接口）并重新排序列表。

到目前为止，这是我的方法：

public static void stringRadixSort(List<String> list, int letters) {
    List<String>[] buckets = (List<String>[]) Array.newInstance(List.class, 26);

    int letterNumber = 1; //Sorts list by 1st letter of each word, then 2nd etc.
    for (int i = 0; i < letters; i++) {
        while (!list.isEmpty()) {
            String word = list.remove(list.first());
            if (word.length() > letters) throw new UnsortableException("The list contains a word that holds more letters than the given maximum number of letters."
                    + "\nMax Letters: " + letters + "\nWord: " + word);
            String letter = word.substring(letterNumber - 1, letterNumber); //EXCEPTION THROWN
            char ch = letter.charAt(0);
            int index = ch - 'a';    //gets index of each letter ('a' = buckets[0], 'z' = buckets[25]
            if (buckets[index] == null) {
                buckets[index] = new LinkedList<String>();
            }
            buckets[index].insertLast(word);
        }

        for (int j = 0; j < buckets.length; j++) {
            if (buckets[j] != null) {
                while (!buckets[j].isEmpty()) {
                    list.insertLast(buckets[j].remove(buckets[j].first()));
                }
            }
        }
        letterNumber++;
    }
}

我的方法的（唯一，我希望）问题是，当我阅读单词的每个字符时，我会创建单词的单个字母子字符串。由于外部for 循环运行letters 次（其中letters 是列表中单词的最大长度），当此循环的迭代大于当前单词的长度时，将引发异常 -即letterNumber > word.length() - 因此它尝试使用大于字符串长度的字符串索引创建子字符串。

如何调整我的方法，使其仅创建每个单词的子字符串，直到 letterNumber == word.length()，然后还能够将排序算法应用于这些较短的单词 - “a”将变为“aa”之前。

Answer 1

你为什么不替换

String letter = word.substring(letterNumber - 1, letterNumber);
char ch = letter.charAt(0);

与

char ch = word.charAt(letterNumber - 1);

直接为您提供char。但这并不能解决IndexOutOfBoundException 的问题。

您当然应该捕获异常并处理它。也许为这种情况创建一个桶是好的：当单词对于当前迭代来说太短时，它被排序到一个桶中。将列表重新合并在一起时，请先获取此存储桶的元素。

public static void stringRadixSort(List<String> list, int letters) {
    List<String>[] buckets = (List<String>[]) Array.newInstance(List.class, 27);

    int letterNumber = 1; //Sorts list by 1st letter of each word, then 2nd etc.
    for (int i = 0; i < letters; i++) {
        while (!list.isEmpty()) {
            String word = list.remove(list.first());
            if (word.length() > letters) throw new UnsortableException("The list contains a word that holds more letters than the given maximum number of letters."
                + "\nMax Letters: " + letters + "\nWord: " + word);
            int index;
            if(word.length() > letterNumber) {
                char ch = word.charAt(letterNumber - 1);
                index = ch - 'a' + 1;    //gets index of each letter ('a' = buckets[1], 'z' = buckets[26], buckets[0] is for short words
            } else {
                index = 0;
            }
            if (buckets[index] == null) {
                buckets[index] = new LinkedList<String>();
            }
            buckets[index].insertLast(word);
        }

        for (int j = 0; j < buckets.length; j++) {
            if (buckets[j] != null) {
                while (!buckets[j].isEmpty()) {
                    list.insertLast(buckets[j].remove(buckets[j].first()));
                }
            }
        }
        letterNumber++;
    }
}

Answer 2

只需将短于字符串长度的元素分组到一个附加组中。您还需要首先对最不重要（相关）的字符进行排序。以下代码使用 java 集合而不是您使用的任何数据结构：

public static void stringRadixSort(List<String> list, int letters) {
    if (list.size() <= 1) {
        return;
    }

    List<String>[] buckets = new List[27];
    for (int i = 0; i < buckets.length; i++) {
        buckets[i] = new LinkedList<>();
    }
    int largestLength = -1;
    int secondLargestLength = 0;
    for (String s : list) {
        int length = s.length();
        if (length >= largestLength) {
            secondLargestLength = largestLength;
            largestLength = length;
        } else if (secondLargestLength < length) {
            secondLargestLength = length;
        }
    }

    if (largestLength > letters) {
        throw new IllegalArgumentException("one of the strings is too long");
    }

    for (int i = secondLargestLength == largestLength ? secondLargestLength-1 : secondLargestLength; i >= 0; i--) {
        for (String word : list) {
            int index = (word.length() <= i) ? 0 : word.charAt(i) - ('a' - 1);
            buckets[index].add(word);
        }

        list.clear();

        for (List<String> lst : buckets) {
            if (lst != null) {
                list.addAll(lst);
                lst.clear();
            }
        }
    }
}

Answer 3

在我所有的尝试中，我一直按照最重要的字母（每个单词的第一个字母）对单词进行排序，然后是下一个重要的字母，依此类推。当然，基数排序依赖于对最低有效位/字母（数字/单词的最后一位/字母）进行排序。因此，我没有从关注letterNumber = 1 开始迭代我的外部for 循环并在每次迭代后递增它，而是从letterNumber = maxWordLength 开始，然后在每次迭代后递减它，以便每次迭代比较下一个最重要的字母。

@SuppressWarnings("unchecked")
public static void stringRadixSort(List<String> list) {
    List<String>[] buckets = (List<String>[]) Array.newInstance(List.class, 27);

    //Find longest word in list
    int maxWordLength = 0;
    for (String word : list) {
        if (word.length() > maxWordLength) {
            maxWordLength = word.length();
        }
    }

    //Sorts list based on least significant letter (last letter of word) to most significant
    int letterNumber = maxWordLength;
    for (int i = 0; i < maxWordLength; i++) {
        while (!list.isEmpty()) {
            String word = list.remove(list.first());
            int index = 0;
            if(word.length() >= letterNumber) {
                char ch = word.charAt(letterNumber - 1);
                index = ch - 'a' + 1;    //gets index of each letter ('a' = buckets[1], 'z' = buckets[26], buckets[0] is for words shorter than 'letterNumber')
            }
            if (buckets[index] == null) {
                buckets[index] = new LinkedList<String>();
            }
            buckets[index].insertLast(word);
        }

        for (int j = 0; j < buckets.length; j++) {
            if (buckets[j] != null) {
                while (!buckets[j].isEmpty()) {
                    list.insertLast(buckets[j].remove(buckets[j].first()));
                }
            }
        }
        letterNumber--;
    }
}