一、字典的实现
dic 可以使用list来实现
i（索引） = hash(key) % solt(槽位数)
此时i重复了怎么办（hash冲突）？
1、拉链法
每个槽位上拉一个List，就是拉链法
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
|
In [6]: solts = []
# 初始化一个list
In [7]: solts_num = 32
# 假设32个槽位
In [8]:
for
_
in
range(solts_num):
...: solts.append([])
...:
In [9]: solts
# 每个槽位上都是一个list
Out[9]:
[[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[]]
## 定义dict的put方法
In [10]: def put(solts, key, value):
...: i =
hash
(key) % solts_num
...: solts[i].append((key, value))
...:
## 定义dict的get方法
In [12]: def get(solts, key):
...: i =
hash
(key) % solts_num
...:
for
k,
v
in
solts[i]:
...:
if
k == key:
...:
return
v
...: raise KeyError(key)
...:
## 测试
In [23]: put(solts,
'a'
, 1)
In [24]: solts
Out[24]:
[[(
'a'
, 1)],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[]]
In [25]: put(solts,
'b'
, 2)
In [26]: put(solts,
'f'
, 9)
In [27]: solts
Out[27]:
[[(
'a'
, 1)],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[(
'f'
, 9)],
[],
[(
'b'
, 2)],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[]]
In [28]: get(solts,
'b'
)
Out[28]: 2
In [29]: get(solts,
'f'
)
Out[29]: 9
In [107]: get(solts,
'x'
)
---------------------------------------------------------------------------
KeyError Traceback (most recent call last)
<ipython-input-107-bdc41302b4b3>
in
<module>()
----> 1 get(solts,
'x'
)
<ipython-input-100-31c1edecb2ae>
in
get(solts, key)
4
if
k == key:
5
return
v
----> 6 raise KeyError(key)
7
KeyError:
'x'
|
此时put一个已存在的key时：
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
|
In [33]: put(solts,
'a'
, 11)
In [34]: solts
Out[34]:
[[(
'a'
, 1), (
'a'
, 11)],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[(
'f'
, 9)],
[],
[(
'b'
, 2)],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[]]
In [35]: get(solts,
'a'
)
Out[35]: 1
|
重新定put方法：
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
|
In [87]: def put(solts, key, value):
...: i =
hash
(key) % solts_num
...:
for
p, (k,
v
)
in
enumerate(solts[i]):
...:
if
k == key:
...: solts[i][p] = (key, value)
...:
else
:
...: solts[i].append((key, value))
In [89]: solts
Out[89]: []
In [90]:
for
_
in
range(solts_num):
...: solts.append([])
...:
In [92]: put(solts,
'a'
, 1)
In [94]: put(solts,
'b'
, 2)
In [95]: put(solts,
'f'
, 8)
In [96]: solts
Out[96]:
[[(
'a'
, 1)],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[(
'f'
, 8)],
[],
[(
'b'
, 2)],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[]]
In [101]: get(solts,
'b'
)
Out[101]: 2
In [102]: get(solts,
'f'
)
Out[102]: 8
In [103]: get(solts,
'a'
)
Out[103]: 11
In [104]: put(solts,
'b'
, 22)
In [105]: get(solts,
'b'
)
Out[105]: 22
In [106]: solts
Out[106]:
[[(
'a'
, 11), (
'a'
, 11)],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[(
'f'
, 8)],
[],
[(
'b'
, 22), (
'b'
, 22)],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[],
[]]
In [107]: get(solts,
'x'
)
---------------------------------------------------------------------------
KeyError Traceback (most recent call last)
<ipython-input-107-bdc41302b4b3>
in
<module>()
----> 1 get(solts,
'x'
)
<ipython-input-100-31c1edecb2ae>
in
get (solts, key)
4
if
k == key:
5
return
v
----> 6 raise KeyError(key)
7
KeyError:
'x'
|
set在底层的实现就是一个忽略了value的dict
2、开地址法
使用某个算法重新计算i，就交开地址法
常用，效率更高，
i = fn(key, i)
待后续学习
本文转自xiexiaojun51CTO博客，原文链接：http://blog.51cto.com/xiexiaojun/1933230 ，如需转载请自行联系原作者
