数据结构与算法 Python语言实现 第三章练习

Posted 2021-03-04 xiaomingyang

巩固

 1 # R-3.2
 2 # n0 = 16
 3 
 4 # R-3.3
 5 # n0 = 20
 6 
 7 # R-3.4
 8 # 常数函数,线性函数?
 9 
10 # R-3.5
11 # logn的c次方 = clogn 斜率是固定常数,故为直线
12 
13 # R-3.6
14 # 2*(0 + 1 + 2 + ... + n) = 2*(n*(n+1))/2 = n*(n+1)
15 
16 # R-3.7
17 # 大O表示法表示的是最坏情况下的算法复杂度
18 
19 # R-3.23
20 O(n)
21 
22 # R-3.24
23 O(n)
24 
25 # R-3.25
26 O(n²)
27 
28 # R-3.26
29 O(n)
30 
31 # R-3.27
32 O(n³)
33 
34 # R-3.29
35 O(nlogn)
36 
37 # R-3.30
38 O(nlogn)
39 
40 # R-3.31
41 最好运行时间: O(nlogn)
42 最坏运行时间: O(n²)
43 
44 # R-3.32
45 O(n)
46 
47 # R-3.33
48 有干扰因素,当数值较小时,干扰因素影响较大
49 
50 # R-3.34
51 1/1 + 1/2 + 1/3 + ... + 1/j + ... + 1/n (n调和数)

巩固部分代码参考

创新

  1 # C-3.35
  2 # 使用排序加遍历相邻三数是否相同: O(nlogn) + O(n) ---> O(nlogn)
  3 
  4 # C-3.36
  5 import random
  6 
  7 
  8 class GetBigNum:
  9     def __init__(self, data):
 10         self.data = data
 11         self.n = len(data)
 12 
 13     def get_some_bigger(self, num):
 14         bigger_data = []
 15         while len(self.data) > self.n - num:
 16             max_val = self.get_max()
 17             bigger_data.append(max_val)
 18             self.data.remove(max_val)
 19         return bigger_data
 20 
 21     def get_max(self):
 22         max_val = self.data[0]
 23         for i in range(1, len(self.data)):
 24             if self.data[i] > max_val:
 25                 max_val = self.data[i]
 26         return max_val
 27 
 28 
 29 # 算法运行时间:O(n)
 30 # test_data = [1, 2, 3, 4, 5, 6, 8, 10, 11, 13, 14, 15, 15, 16, 17, 18, 19]
 31 # big_num = GetBigNum(test_data)
 32 # print(big_num.get_some_bigger(10))
 33 
 34 # 或者使用排序加data[-10:]  O(nlogn)
 35 
 36 # C-3.45
 37 # def find_unique_lost(s):
 38 #     total = 1
 39 #     for i in range(len(s) + 1):
 40 #         total *= (i + 1)
 41 #     for j in s:
 42 #         total /= (j + 1)
 43 #     return int(total - 1)
 44 
 45 
 46 # test_s = [0, 1, 2, 3, 4, 6]
 47 # print(find_unique_lost(test_s))
 48 
 49 
 50 # C-3.46
 51 # 2只绵羊是特例
 52 
 53 
 54 # C-3.50
 55 # 1.
 56 def sum_num(x, n, data):
 57     sum_result = 0
 58 
 59     for i in range(n+1):
 60         sum_xi = 1
 61         for j in range(i):
 62             sum_xi *= x
 63         sum_result += data[i] * sum_xi
 64     return sum_result
 65 
 66 
 67 # print(sum_num(2, 3, [1, 2, 3, 4]))
 68 
 69 
 70 # 2.
 71 def sum_num2(x, n, data):
 72     sum_result = 0
 73     sum_xi = 1
 74     for i in range(n+1):
 75         sum_xi *= x
 76         sum_result += data[i] * sum_xi
 77     return sum_result / x
 78 
 79 
 80 # print(sum_num2(2, 3, [1, 2, 3, 4]))
 81 
 82 
 83 # C-3.54
 84 from collections import defaultdict
 85 
 86 def get_max_amount(data):
 87     total_amount = 0
 88     n = len(data)
 89     max_amount = 0
 90     max_amount_dic = defaultdict()
 91     max_amount_dic[‘max_amount‘] = dict()
 92     while total_amount < n:
 93         for i in range(n):
 94             val_amount = 1
 95             for val in data[i+1:]:
 96                 if data[i] == val:
 97                     val_amount += 1
 98             if val_amount >= max_amount:
 99                 max_amount = val_amount
100                 max_amount_val = set()
101                 max_amount_val.add(data[i])
102                 if max_amount in max_amount_dic[‘max_amount‘]:
103                     max_amount_val.update(max_amount_dic[‘max_amount‘])
104                 max_amount_dic[‘max_amount‘] = {max_amount: max_amount_val}
105                 total_amount += val_amount
106     return max_amount_dic
107 
108 
109 # test_data = [1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 60]
110 # print(get_max_amount(test_data))
111 #
112 # S = []
113 # for i in range(1000):
114 #     S.append(random.randint(0, 4000))
115 # print(get_max_amount(S))
116 
117 
118 # 更优解
119 def find_most_frequent(n):
120     s = []
121     for _ in range(n):
122         s.append(random.randint(0, 4*n - 1))
123 
124     restore_list = [0] * (4*n)
125     most_int = 0
126     for num in s:
127         restore_list[num] += 1
128         if restore_list[num] >= restore_list[most_int]:
129             most_int = num
130     if most_int == 1:
131         return False
132     else:
133         return most_int, restore_list[most_int]
134 
135 
136 print(find_most_frequent(1000))

创新部分代码参考