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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.f
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.
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 };
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原文地址：http://www.cnblogs.com/lishiblog/p/4183784.html
