python实现散列表的链表法

　　在散列中，链接法是一种最简单的碰撞解决技术，这种方法的原理就是把散列到同一槽中的所有元素

都放在一个链表中。

　　链接法有两个定理，定理一：

　　在简单一致散列的假设下，一次不成功查找的期望时间为O(1 + n)

　　定理二：

　　在简单一致散列的假设下，平均情况下一次成功的查找需要的时间为O（1 + n）

　　该方法使用python的实现如下：

# keys函数
#!/usr/bin/env python
# coding=utf-8

class Dict:
def __init__(self, num):
self.__solts__ = []
self.num = num
for _ in range(num):
self.__solts__.append([])
def hash_fun(self,key,num):
hashval = 0
x = key
if x < 0:
print "the key is low"
return
while x != 0:
hashval = (hashval << 3) + x%10
x /=10
return hashval % num
def put(self, key, value):
i = self.hash_fun(key,self.num) % self.num
for p, (k, v) in enumerate(self.__solts__[i]):
if k == key:
break
else:
self.__solts__[i].append((key, value))
return
self.__solts__[i][p] = (key, value)
def get(self, key):
i = self.hash_fun(key,self.num) % self.num
for k, v in self.__solts__[i]:
if k == key:
return v
raise KeyError(key)
# keys函数
def keys(self):
ret = []
for solt in self.__solts__:
for k, _ in solt:
ret.append(k)
return ret
def __getitem__(self,key):
return self.get(key)

def __setitem__(self,key,data):
self.put(key,data)
H = Dict(13)
H[54] = "cat"
H[26] = "dog"
H[93] = "lion"
H[17] = "tiger"
H[77] = "bird"
H[31] = "cow"
H[44] = "goat"
H[55] = "pig"
H[20] = "chicken"
print H.get(54)
H.put(13,"duck")
print H.__solts__
print H.keys()

　　运行结果：

cat
[[], [], [(20, 'chicken')], [], [], [(17, 'tiger')], [(55, 'pig')], [(93, 'lion')], [], [], [(44, 'goat')], [(54, 'cat'), (26, 'dog'), (77, 'bird'), (31, 'cow')], [(13, 'duck')]]
[20, 17, 55, 93, 44, 54, 26, 77, 31, 13]

　　参考文档：算法导论