java 中HashCode重复的可能性

public int hashCode() { 
  int h = hash; 
  if (h == 0) { 
    int off = offset; 
    char val[] = value; 
    int len = count; 
 
      for (int i = 0; i < len; i++) { 
        h = 31*h + val[off++]; 
      } 
      hash = h; 
    } 
    return h; 
  }

import java.util.HashMap; 
 
public class Test { 
 
  static HashMap map = new HashMap(); 
 
  private static char startChar = 'A'; 
 
  private static char endChar = 'z'; 
 
  private static int offset = endChar - startChar + 1; 
 
  private static int dup = 0; 
 
  public static void main(String[] args) { 
    int len = 3; 
    char[] chars = new char[len]; 
    tryBit(chars, len); 
    System.out.println((int)Math.pow(offset, len) + ":" + dup); 
  } 
 
  private static void tryBit(char[] chars, int i) { 
    for (char j = startChar; j <= endChar; j++) { 
      chars[i - 1] = j; 
      if (i > 1) 
        tryBit(chars, i - 1); 
      else 
        test(chars); 
    } 
  } 
 
  private static void test(char[] chars) { 
 
    String str = new String(chars).replaceAll("[^a-zA-Z_]", "").toUpperCase();// 195112:0 
    //String str = new String(chars).toLowerCase();//195112:6612 
    //String str = new String(chars).replaceAll("[^a-zA-Z_]","");//195112:122500 
    //String str = new String(chars);//195112:138510 
    int hash = str.hashCode(); 
    if (map.containsKey(hash)) { 
      String s = (String) map.get(hash); 
      if (!s.equals(str)) { 
        dup++; 
        System.out.println(s + ":" + str); 
      } 
    } else { 
      map.put(hash, str); 
      // System.out.println(str); 
    } 
  } 
}