後綴數組實現錯誤

Question

我一直收到編譯器錯誤，執行後綴數組Arrays.sort。

我收到以下錯誤：

一個不能被解析爲一個變量
語法錯誤令牌「」。預計
語法錯誤令牌「 - 」， - 預期
一個不能被解析爲一個變量
B不能被解析爲一個變量

在下面的代碼：

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

a無法解析爲變量
有關令牌「，」，「」的語法錯誤。預計
語法錯誤令牌「 - 」， - 預期
一個不能被解析爲一個變量
B不能被解析爲一個變量

你在這一行遇到這些錯誤（出現這在兩次的代碼）：

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

原因必須是，你是不是在編譯的Java 8 Lambda表達式的代碼需要java 8.