Question

我根据这个site使用了Boyer Moore算法。这仅在文本中实现模式搜索一次，程序退出。任何人都可以帮我修改这段代码，以便用它们的起始和结束索引多次找到该模式吗？

    public class BoyerMoore {
        private final int R;     // the radix
        private int[] right;     // the bad-character skip array
        private String pat;      // or as a string

        // pattern provided as a string
        public BoyerMoore(String pat) {
            this.R = 256;
            this.pat = pat;

            // position of rightmost occurrence of c in the pattern
            right = new int[R];
            for (int c = 0; c < R; c++)
                right[c] = -1;
            for (int j = 0; j < pat.length(); j++)
                right[pat.charAt(j)] = j;
        }

        // return offset of first match; N if no match
        public ArrayList<Integer> search(String txt) {
            int M = pat.length();
            int N = txt.length();
            ArrayList<Integer> newArrayInt = new ArrayList<Integer>();
            int skip;
            for (int i = 0; i <= N - M; i += skip) {
                skip = 0;
                for (int j = M-1; j >= 0; j--) {
                    if (pat.charAt(j) != txt.charAt(i+j)) {
                        skip = Math.max(1, j - right[txt.charAt(i+j)]);
                        break;
                    }
                }
                if (skip == 0) 
                    newArrayInt.add(i);    // found
            }
            return newArrayInt;                       // not found
        }

        // test client
        public static void main(String[] args) {
            String pat = "abc";
            String txt = "asdf ghjk klll abc qwerty abc and poaslf abc";

            BoyerMoore boyermoore1 = new BoyerMoore(pat);

            ArrayList<Integer> offset = boyermoore1.search(txt);

            // print results
            System.out.println("Offset: "+ offset);


 }
}

Answer 1

我明白了。当它在文本中找到模式时，跳过始终为0。

public class BoyerMoore {
    private final int R;     // the radix
    private int[] right;     // the bad-character skip array
    private String pat;      // or as a string

    // pattern provided as a string
    public BoyerMoore(String pat) {
        this.R = 256;
        this.pat = pat;

        // position of rightmost occurrence of c in the pattern
        right = new int[R];
        for (int c = 0; c < R; c++)
            right[c] = -1;
        for (int j = 0; j < pat.length(); j++)
            right[pat.charAt(j)] = j;
    }

    // return offset of first match; N if no match
    public ArrayList<Integer> search(String txt) {
        int M = pat.length();
        int N = txt.length();
        ArrayList<Integer> newArrayInt = new ArrayList<Integer>();
        int skip;
        for (int i = 0; i <= N - M; i += skip) {
            skip = 0;
            for (int j = M-1; j >= 0; j--) {
                if (pat.charAt(j) != txt.charAt(i+j)) {
                    skip = Math.max(1, j - right[txt.charAt(i+j)]);
                    break;
                }
            }
            if (skip == 0) 
            {
                newArrayInt.add(i);    // found
                skip++;
            }
        }
        return newArrayInt;                       // not found
    }

    // test client
    public static void main(String[] args) {
        String pat = "abc";
        String txt = "asdf ghjk klll abc qwerty abc and poaslf abc";

        BoyerMoore boyermoore1 = new BoyerMoore(pat);

        ArrayList<Integer> offset = boyermoore1.search(txt);

        // print results
        System.out.println("Offset: "+ offset);
    }
}

Answer 2

public class Boyer {
    /**
     * @param args
     */
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        Scanner get= new Scanner(System.in);
      String m,n;
      int i,j;
      String T,P;
      T=get.nextLine();
      System.out.println("Text T is"+T);
      P=get.nextLine();
      System.out.println("Pattern P is"+P);
      int n1=T.length();
      int m1=P.length();
      for(i=0;i<=n1-m1;i++){
           j=0;
          while(j<m1 && (T.charAt(i+j)==P.charAt(j))){
               j=j+1;
               if(j==m1)
                   System.out.println("match found at"+i);
          }
      }
    }
}

修改Boyer Moore以进行多种模式搜索

2 个答案: