• LintCode-Hash Function


    In data structure Hash, hash function is used to convert a string(or any other type) into an integer smaller than hash size and bigger or equal to zero. The objective of designing a hash function is to "hash" the key as unreasonable as possible. A good hash function can avoid collision as less as possible. A widely used hash function algorithm is using a magic number 33, consider any string as a 33 based big integer like follow:

    hashcode("abcd") = (ascii(a) * 333ascii(b) * 332ascii(c) *33 + ascii(d)) % HASH_SIZE 

                                  = (97* 333 + 98 * 332 + 99 * 33 +100) % HASH_SIZE

                                  = 3595978 % HASH_SIZE

    here HASH_SIZE is the capacity of the hash table (you can assume a hash table is like an array with index 0 ~ HASH_SIZE-1).

    Given a string as a key and the size of hash table, return the hash value of this key.f

    Example

    For key="abcd" and size=100, return 78

    Clarification

    For this problem, you are not necessary to design your own hash algorithm or consider any collision issue, you just need to implement the algorithm as described.

    Analysis:

    We need to be careful about overflow, when we calculate the intermiedate result, we need be careful with overflow.

    Solution 1:

    Use long type.

     1 class Solution {
     2     /**
     3      * @param key: A String you should hash
     4      * @param HASH_SIZE: An integer
     5      * @return an integer
     6      */
     7     public int hashCode(char[] key,int HASH_SIZE) {
     8         if (key.length==0) return 0;
     9         int res = 0;
    10         int base = 1;
    11         for (int i=key.length-1;i>=0;i--){
    12             res += modMultiply((int)key[i],base,HASH_SIZE);;
    13             res %= HASH_SIZE;
    14             base = modMultiply(base,33,HASH_SIZE);
    15         }
    16         return res;
    17     }
    18 
    19     public int modMultiply(long a, long b, int HASH_SIZE){        
    20         long temp = a*b%HASH_SIZE;
    21         return (int) temp;
    22     }
    23         
    24 };

    Solution 2:

    Use int type to perform the multiply. However, change the way to calculate the whole expression. Using the way used in solution 1 will cause TLE.

     1 class Solution {
     2     /**
     3      * @param key: A String you should hash
     4      * @param HASH_SIZE: An integer
     5      * @return an integer
     6      */
     7     public int hashCode(char[] key,int HASH_SIZE) {
     8         if (key.length==0) return 0;
     9         int res = 0;
    10         int base = 33;
    11         for (int i=0;i<key.length;i++){
    12             res = modMultiply(res,base,HASH_SIZE);
    13             res += key[i];
    14             res = res % HASH_SIZE;
    15         }
    16         return res;
    17     }
    18 
    19
    20     public int modMultiply(int a, int b, int HASH_SIZE){        
    21         int res = a;
    22         for (int j=1;j<b;j++){
    23             int temp = (a-HASH_SIZE);
    24             if (res+temp>=0)     res = res+temp;
    25             else res = res + a;
    26         }
    27         return res;
    28     }
    29         
    30 };
  • 相关阅读:
    informix 外部表 pipe
    关于XML的一些解析操作
    oracle 导出导入数据库
    判断请求访问的浏览器类型设备
    git与SVN的区别
    Java获取文件路径
    <DIV>内容显示隐藏功能实现
    文件下载
    文件上传
    记录启动Nginx启动失败
  • 原文地址:https://www.cnblogs.com/lishiblog/p/4183784.html
Copyright © 2020-2023  润新知