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);     }   } }