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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) * 333 + ascii(b) * 332 + ascii(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.
For key="abcd" and size=100, return 78
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.
这道题要注意的就是溢出,要注意两点:
1. 使用long类型
2. 每次计算33的n次方要再做%HASH_SIZE,每一项累加之后也要%HASH_SIZE避免溢出
class Solution { /** * @param key: A String you should hash * @param HASH_SIZE: An integer * @return an integer */ public int hashCode(char[] key,int HASH_SIZE) { // write your code here if(key == null || key.length == 0) return 0; long factor = 1; long result = 0; for(int i = key.length - 1; i >= 0; i--){ result += (((int) key[i]) * factor) % HASH_SIZE; factor = (factor * 33) % HASH_SIZE; } result %= HASH_SIZE; return (int) result; } };
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原文地址:http://www.cnblogs.com/goblinengineer/p/5215685.html