上次大致分析了一下哈希表的链地址法的实现,今天来分析一下另一种解决哈希冲突的做法,即为每个Hash值,建立一个Hash桶(Bucket),桶的容量是固定的,也就是只能处理固定次数的冲突,如1048576个Hash桶,每个桶中有4个表项(Entry),总计4M个表项。其实这两种的实现思路雷同,就是对Hash表中每个Hash值建立一个冲突表,即将冲突的几个记录以表的形式存储在其中;
废话不多说,上代码和图示基本能说明清楚:
完整的代码,请看:这里,一位圣安德鲁斯大学的讲师:KRISTENSSON博客
这里截取几个主要的片段:
主要的数据结构:
struct Pair { char *key; char *value; }; struct Bucket { unsigned int count; Pair *pairs; }; struct StrMap { unsigned int count; Bucket *buckets; };
主要的函数:
put:
int sm_put(StrMap *map, const char *key, const char *value) { unsigned int key_len, value_len, index; Bucket *bucket; Pair *tmp_pairs, *pair; char *tmp_value; char *new_key, *new_value; if (map == NULL) { return 0; } if (key == NULL || value == NULL) { return 0; } key_len = strlen(key); value_len = strlen(value); /* Get a pointer to the bucket the key string hashes to */ index = hash(key) % map->count; bucket = &(map->buckets[index]); /* Check if we can handle insertion by simply replacing * an existing value in a key-value pair in the bucket. */ if ((pair = get_pair(bucket, key)) != NULL) { /* The bucket contains a pair that matches the provided key, * change the value for that pair to the new value. */ if (strlen(pair->value) < value_len) { /* If the new value is larger than the old value, re-allocate * space for the new larger value. */ tmp_value = realloc(pair->value, (value_len + 1) * sizeof(char)); if (tmp_value == NULL) { return 0; } pair->value = tmp_value; } /* Copy the new value into the pair that matches the key */ strcpy(pair->value, value); return 1; } /* Allocate space for a new key and value */ new_key = malloc((key_len + 1) * sizeof(char)); if (new_key == NULL) { return 0; } new_value = malloc((value_len + 1) * sizeof(char)); if (new_value == NULL) { free(new_key); return 0; } /* Create a key-value pair */ if (bucket->count == 0) { /* The bucket is empty, lazily allocate space for a single * key-value pair. */ bucket->pairs = malloc(sizeof(Pair)); if (bucket->pairs == NULL) { free(new_key); free(new_value); return 0; } bucket->count = 1; } else { /* The bucket wasn't empty but no pair existed that matches the provided * key, so create a new key-value pair. */ tmp_pairs = realloc(bucket->pairs, (bucket->count + 1) * sizeof(Pair)); if (tmp_pairs == NULL) { free(new_key); free(new_value); return 0; } bucket->pairs = tmp_pairs; bucket->count++; } /* Get the last pair in the chain for the bucket */ pair = &(bucket->pairs[bucket->count - 1]); pair->key = new_key; pair->value = new_value; /* Copy the key and its value into the key-value pair */ strcpy(pair->key, key); strcpy(pair->value, value); return 1; }
get:
int sm_get(const StrMap *map, const char *key, char *out_buf, unsigned int n_out_buf) { unsigned int index; Bucket *bucket; Pair *pair; if (map == NULL) { return 0; } if (key == NULL) { return 0; } index = hash(key) % map->count; bucket = &(map->buckets[index]); pair = get_pair(bucket, key); if (pair == NULL) { return 0; } if (out_buf == NULL && n_out_buf == 0) { return strlen(pair->value) + 1; } if (out_buf == NULL) { return 0; } if (strlen(pair->value) >= n_out_buf) { return 0; } strcpy(out_buf, pair->value); return 1; }
哈希函数:
/* * Returns a hash code for the provided string. */ static unsigned long hash(const char *str) { unsigned long hash = 5381; int c; while (c = *str++) { hash = ((hash << 5) + hash) + c; } return hash; }
大致的思路是这样的:
首先哈希桶的个数是固定的,有用户构建的时候输入,一旦构建,个数就已经固定;查找的时候首先将key值通过哈希函数获取哈希值,根据哈希值获取到对应的哈希桶,然后遍历哈希桶内的pairs数组获取;
这两种实现方法看似比较类似,但也有差异:
基于哈希桶的情况下,由于Hash桶容量的限制,所以,有可能发生Hash表填不满的情况,也就是,虽然Hash表里面还有空位,但是新建的表项由于冲突过多,而不能装入Hash表中。不过,这样的实现也有其好处,就是查表的最大开销是可以确定的,因为最多处理的冲突数是确定的,所以算法的时间复杂度为O(1)+O(m),其中m为Hash桶容量。
而另一种通过链表的实现,由于Hash桶的容量是无限的,因此,只要没有超出Hash表的最大容量,就能够容纳新建的表项。但是,一旦发生了Hash冲突严重的情况,就会造成Hash桶的链表过长,大大降低查找效率。在最坏的情况下,时间复杂度退化为O(n),其中n为Hash表的总容量。当然,这种情况的概率小之又小,几乎是可以忽略的。