后綴數組實現錯誤

Question

我通過Arrays.sort的后綴數組實現不斷收到編譯器錯誤。

我收到以下錯誤：

一個不能解析為變量
令牌“，” 、.的語法錯誤。 預期
令牌“-”的語法錯誤，-預期
一個不能解析為變量
b無法解析為變量

在下面的代碼中：

import java.util.*;

public class SuffixArray {

  // sort suffixes of S in O(n*log(n))
  public static int[] suffixArray(CharSequence S) {
    int n = S.length();
    Integer[] order = new Integer[n];
    for (int i = 0; i < n; i++)
      order[i] = n - 1 - i;

    // stable sort of characters
    Arrays.sort(order, (a, b) -> Character.compare(S.charAt(a), S.charAt(b)));

    int[] sa = new int[n];
    int[] classes = new int[n];
    for (int i = 0; i < n; i++) {
      sa[i] = order[i];
      classes[i] = S.charAt(i);
    }
    // sa[i] - suffix on i'th position after sorting by first len characters
    // classes[i] - equivalence class of the i'th suffix after sorting by first len characters

    for (int len = 1; len < n; len *= 2) {
      int[] c = classes.clone();
      for (int i = 0; i < n; i++) {
        // condition sa[i - 1] + len < n simulates 0-symbol at the end of the string
        // a separate class is created for each suffix followed by simulated 0-symbol
        classes[sa[i]] = i > 0 && c[sa[i - 1]] == c[sa[i]] && sa[i - 1] + len < n && c[sa[i - 1] + len / 2] == c[sa[i] + len / 2] ? classes[sa[i - 1]] : i;
      }
      // Suffixes are already sorted by first len characters
      // Now sort suffixes by first len * 2 characters
      int[] cnt = new int[n];
      for (int i = 0; i < n; i++)
        cnt[i] = i;
      int[] s = sa.clone();
      for (int i = 0; i < n; i++) {
        // s[i] - order of suffixes sorted by first len characters
        // (s[i] - len) - order of suffixes sorted only by second len characters
        int s1 = s[i] - len;
        // sort only suffixes of length > len, others are already sorted
        if (s1 >= 0)
          sa[cnt[classes[s1]]++] = s1;
      }
    }
    return sa;
  }

  // sort rotations of S in O(n*log(n))
  public static int[] rotationArray(CharSequence S) {
    int n = S.length();
    Integer[] order = new Integer[n];
    for (int i = 0; i < n; i++)
      order[i] = i;
    Arrays.sort(order, (a, b) -> Character.compare(S.charAt(a), S.charAt(b)));
    int[] sa = new int[n];
    int[] classes = new int[n];
    for (int i = 0; i < n; i++) {
      sa[i] = order[i];
      classes[i] = S.charAt(i);
    }
    for (int len = 1; len < n; len *= 2) {
      int[] c = classes.clone();
      for (int i = 0; i < n; i++)
        classes[sa[i]] = i > 0 && c[sa[i - 1]] == c[sa[i]] && c[(sa[i - 1] + len / 2) % n] == c[(sa[i] + len / 2) % n] ? classes[sa[i - 1]] : i;
      int[] cnt = new int[n];
      for (int i = 0; i < n; i++)
        cnt[i] = i;
      int[] s = sa.clone();
      for (int i = 0; i < n; i++) {
        int s1 = (s[i] - len + n) % n;
        sa[cnt[classes[s1]]++] = s1;
      }
    }
    return sa;
  }

  // longest common prefixes array in O(n)
  public static int[] lcp(int[] sa, CharSequence s) {
    int n = sa.length;
    int[] rank = new int[n];
    for (int i = 0; i < n; i++)
      rank[sa[i]] = i;
    int[] lcp = new int[n - 1];
    for (int i = 0, h = 0; i < n; i++) {
      if (rank[i] < n - 1) {
        for (int j = sa[rank[i] + 1]; Math.max(i, j) + h < s.length() && s.charAt(i + h) == s.charAt(j + h); ++h)
      ;
        lcp[rank[i]] = h;
        if (h > 0)
          --h;
      }    
    }
    return lcp;
  }

  // Usage example
  public static void main(String[] args) {
    String s1 = "abcab";
    int[] sa1 = suffixArray(s1);

    // print suffixes in lexicographic order
    for (int p : sa1)
      System.out.println(s1.substring(p));

    System.out.println("lcp = " + Arrays.toString(lcp(sa1, s1)));

    // random test
    Random rnd = new Random(1);
    for (int step = 0; step < 100000; step++) {
      int n = rnd.nextInt(100) + 1;
      StringBuilder s = new StringBuilder();
      for (int i = 0; i < n; i++)
        s.append((char) ('\1' + rnd.nextInt(10)));
      int[] sa = suffixArray(s);
      int[] ra = rotationArray(s.toString() + '\0');
      int[] lcp = lcp(sa, s);
  for (int i = 0; i + 1 < n; i++) {
        String a = s.substring(sa[i]);
        String b = s.substring(sa[i + 1]);
        if (a.compareTo(b) >= 0
            || !a.substring(0, lcp[i]).equals(b.substring(0, lcp[i]))
            || (a + " ").charAt(lcp[i]) == (b + " ").charAt(lcp[i])
            || sa[i] != ra[i + 1])
          throw new RuntimeException();
      }
    }
    System.out.println("Test passed");
  }
}

Answer 1

一個不能解析為變量
令牌“，” 、.的語法錯誤。 預期
令牌“-”的語法錯誤，-預期
一個不能解析為變量
b無法解析為變量

您在此行上收到這些錯誤（在代碼中出現兩次）：

Arrays.sort(order, (a, b) -> Character.compare(S.charAt(a), S.charAt(b)));
                    ^^    ^                             ^            ^

原因必須是您沒有在Java 8中編譯代碼。Lambda表達式需要Java 8。