Hash Function

Source

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

Have you met this question in a real interview? Yes
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.

 

题解1

基本实现题，大多数人看到题目的直觉是按照定义来递推不就得了嘛，但其实这里面大有玄机，因为在字符串较长时使用 long 型来计算33的幂会溢出！所以这道题的关键在于如何处理大整数溢出。对于整数求模，(a * b) % m = a % m * b % m 这个基本公式务必牢记。根据这个公式我们可以大大降低时间复杂度和规避溢出。

Java

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) {
        if (key == null || key.length == 0) return -1;

        long hashSum = 0;
        for (int i = 0; i < key.length; i++) {
            hashSum += key[i] * modPow(33, key.length - i - 1, HASH_SIZE);
            hashSum %= HASH_SIZE;
        }

        return (int)hashSum;
    }

    private long modPow(int base, int n, int mod) {
        if (n == 0) {
            return 1;
        } else if (n == 1) {
            return base % mod;
        } else if (n % 2 == 0) {
            long temp = modPow(base, n / 2, mod);
            return (temp % mod) * (temp % mod) % mod;
        } else {
            return (base % mod) * modPow(base, n - 1, mod) % mod;
        }
    }
}

源码分析

题解1属于较为直观的解法，只不过在计算33的幂时使用了私有方法modPow, 这个方法使用了对数级别复杂度的算法，可防止 TLE (Time Limit Exceeded) 的产生。注意两个 int 型数据在相乘时可能会溢出，故对中间结果的存储需要使用 long.

复杂度分析

遍历加求modPow，时间复杂度 O(nlogn), 空间复杂度 O(1). 当然也可以使用哈希表的方法将幂求模的结果保存起来，这样一来空间复杂度就是 O(n), 不过时间复杂度为 O(n).

题解2 - 巧用求模公式

从题解1中我们可以看到其时间复杂度还是比较高的，作为基本库来使用是比较低效的。我们从范例hashcode("abc")为例进行说明。

hashcode(abc)=(a×33^2+b×33+c)%M=(33(33×a+b)+c)%M=(33(33(33×0+a)+b)+c)%M

再根据 (a×b)%M=(a%M)×(b%M)

从中可以看出使用迭代的方法较容易实现。

Java

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) {
        if (key == null || key.length == 0) return -1;

        long hashSum = 0;
        for (int i = 0; i < key.length; i++) {
            hashSum = 33 * hashSum + key[i];
            hashSum %= HASH_SIZE;
        }

        return (int)hashSum;
    }
}

源码分析

精华在hashSum = 33 * hashSum + key[i];

复杂度分析

时间复杂度 O(n), 空间复杂度 O(1).