java - 如何在不输入固定长度的情况下使用LSD String Sort？

Question

抱歉我的英语不好。我正在设计 LSD 字符串排序算法，我有一个与之相关的问题。这是我的代码。我希望输入 W not fixed，例如:

String[] a = {"38A", "3TW723", "2IYEA938", "3CI34780720"};

public static void sort(String[] a, int w) {  // Sort a[] on leading W characters.
        int R = 256;
        int N = a.length;
        //For each of the character from right to left
        for (int d = w - 1; d >= 0; d--) {
            //1. count the frequencies
            int[] count = new int[R + 1];
            for (int i = 0; i < N; i++) {
                count[a[i].charAt(d) + 1]++;
            }
            //2. Transform counts to indices
            for (int r = 0; r < R; r++) {
                count[r + 1] += count[r];
            }
            //3. Distribute
            String aux[] = new String[N];
            for (int i = 0; i < N; i++) {
                aux[count[a[i].charAt(d)]] = a[i];
                count[a[i].charAt(d)]++;
            }
            //4. Copyback
            System.arraycopy(aux, 0, a, 0, N);
        }
    }

Answer 1

要开发适用于可变长度字符串的 LSD 字符串排序的实现，我们需要在您的代码的基础上做很多工作。我们需要找到 string[] a 中最长的字符串长度，所以当 d >= a[i].length() 时，我们返回 0，这意味着我们添加额外的 0 使每个字符串的长度相同。这是我的代码。

// Develop an implementation of LSD string sort 
// that works for variable-length strings.
public class LSDForVariableLengthStrings {

    // do not instantiate
    private LSDForVariableLengthStrings() { }

    // find longest length string in string[] a.
    public static int findLongestLength(String[] a) {
        int longest = 0;
        for (int i = 0; i < a.length; ++i) {
            if (a[i].length() > longest) {
                longest = a[i].length();
            }
        }
        return longest;
    }

    // if d >= 0 && d < a[i].length(), return a[i].charAt(d);
    // else , return 0, which means least value to sort.
    public static int findCharAtInString(int i, int d, String[] a) {
        if (d < 0 || d >= a[i].length()) {
            return 0;
        }
        return a[i].charAt(d);
    }

    // Rearranges the array of variable-length strings.
    public static void sort(String[] a) {
        int n = a.length;
        int R = 256;    // extended ASCII alphabet size.
        String[] aux = new String[n];
        int w = findLongestLength(a);  // w is the length of longest string in a.
        for (int d = w - 1; d >= 0; d--) {
            // sort by key-indexed counting on dth character

            // compute frequency counts
            int[] count = new int[R + 1];
            for (int i = 0; i < n; ++i) {
                int c = findCharAtInString(i, d, a);
                count[c + 1]++;
            }

            // compute cumulates
            for (int r = 0; r < R; ++r) {
                count[r + 1] += count[r];
            }

            // move data
            for (int i = 0; i < n; ++i) {
                int c = findCharAtInString(i, d, a);
                aux[count[c]++] = a[i];
            }

            // copy back
            for (int i = 0; i < n; ++i) {
                a[i] = aux[i];
            }
        }
    }
    public static void main(String[] args) {

        String[] a = {"38A", "3TW723", "2IYEA938", "3CI34780720"};
        int n = a.length;
        // sort the strings
        sort(a);

        // prints results
        for (int i = 0; i < n; ++i) {
            System.out.println(a[i]);
        }

    }
}