^_^y

時間複雜度：O(mn)

public class BruteForceAlgorithm {
 public static final String T = "ABABAABBCABC";
 public static final String P = "ABC";
 
 public static void match(char[] t, char[] p)
 {
  for (int i = 0; i < t.length; i++)
  {
   char[] tmp = new char[p.length];
   int index = 0;
   for (int j = 0; j < p.length; j++)
   {
    if (i+j < t.length && t[i+j] == p[j])
    {
     tmp[index++] = p[j];
    }
    else
    {
     continue;
    }
   }
   
   if (index == p.length)
   {
    System.out.println(new String(tmp));
   }
  }
 }
 
 public static void main(String[] args) {
  match(T.toCharArray(), P.toCharArray());
 }
}

void BF(char *x, int m, char *y, int n) {
   int i, j;

   /* Searching */
   for (j = 0; j <= n - m; ++j) {
      for (i = 0; i < m && x[i] == y[i + j]; ++i);
      if (i >= m)
         OUTPUT(j);
   }
}

設有三組集合，在以下每組中取出一個數
{A,B,C}
{D,E,F}
{G,H,I}
可以得到27組 {A,D,G}{A,D,H}{A,D,I}{A,E,G}{A,E,H}{A,E,I}....

public class Permutation
{
    static int index = 1;
    static int MAX = 3; // 用來產生 3*3, 改成4就是4*4, 5*5 6*6....
    static char[] soultion = new char[MAX]; // 用來紀錄排列解.
    static char[][] s = new char[MAX][MAX]; // 陣列.
    static boolean[][] used = new boolean[MAX][MAX]; // 紀錄這個數字是否被使用.
   
    static void perm(int k, int h, int n)
    {
        if (k == n)
        {
            System.out.print("序號" + (index++) + ":");
            for (char s : soultion)
            {
                System.out.print(s + "\t");
            }
            System.out.println();
        }
        else
        {
            for (int i = 0; i < n; i++)
            {
                if (!used[h][i])
                {
                    used[h][i] = true;
                    soultion[k] = s[h][i];
                    perm(k+1, h+1, n);
                    used[h][i] = false;

                }
            }
        }
    }

    public static void main(String[] args)
    {
        // 純粹產生排列陣列字元.
        System.out.println("陣列:");
        int c = 65;
        for (int i = 0; i < MAX; i++)
        {
            for (int j = 0; j < MAX; j++)
            {
                used[i][j] = false;
                s[i][j] = (char) (c++);
            }
        }
        // Print 字元.
        for(char[] ss : s)
        {
            for (char ch : ss)
            {
                System.out.print(ch);
            }
            System.out.println();
        }
       
        perm(0, 0,MAX);
    }
}