集合类型：算法中的数据容器

nums = [10, 20, 30]
nums[0]                 # 10
nums[-1]                # 30
nums[1:3]               # [20, 30]，新列表
nums[::-1]              # 反转后的新列表

nums.append(40)         # 末尾添加，均摊 O(1)
nums.extend([50, 60])   # 批量添加
last = nums.pop()       # 删除末尾，O(1)
first = nums.pop(0)     # 删除开头，O(n)
nums.insert(1, 99)      # 中间插入，O(n)
nums.remove(30)         # 查找并删除第一个 30，O(n)
30 in nums              # 存在性检查，O(n)

a = [1, 2]
b = a                   # b 与 a 指向同一个列表
b.append(3)             # a 也变成 [1, 2, 3]

c = a[:]                # 新的外层列表
d = list(a)

wrong = [[0] * 3] * 4
wrong[0][0] = 1         # 四行的第一个元素都会变成 1

correct = [[0] * 3 for _ in range(4)]

squares = [x * x for x in range(5)]
evens = [x for x in nums if x % 2 == 0]
matrix = [[row * cols + col for col in range(cols)] for row in range(rows)]

point = (3, 5)
x, y = point

single = (7,)           # 单元素元组必须有逗号

visited = {(0, 0)}
visited.add((1, 2))

scores = {"Alice": 95, "Bob": 88}
scores["Alice"]               # key 不存在会抛 KeyError
scores.get("Carol", 0)        # 不存在时返回默认值
scores["Carol"] = 90

for key in scores:
    print(key)

for key, value in scores.items():
    print(key, value)

for value in scores.values():
    print(value)

# 计数
frequency = {}
for value in nums:
    frequency[value] = frequency.get(value, 0) + 1

# 分组
groups = {}
for word in words:
    key = "".join(sorted(word))
    groups.setdefault(key, []).append(word)

# 记录最后出现位置
last_position = {}
for index, value in enumerate(nums):
    last_position[value] = index

seen = set()
seen.add(3)
seen.add(3)             # 仍然只有一个 3
3 in seen               # 平均 O(1)
seen.discard(9)         # 不存在也不会报错

a = {1, 2, 3}
b = {2, 3, 4}

a | b                   # 并集 {1, 2, 3, 4}
a & b                   # 交集 {2, 3}
a - b                   # 差集 {1}
a ^ b                   # 对称差 {1, 4}
a <= b                  # 是否为子集

from collections import deque

queue = deque([1, 2])
queue.append(3)         # 右端加入
queue.appendleft(0)     # 左端加入
queue.pop()             # 右端删除
queue.popleft()         # 左端删除

from collections import Counter

count = Counter("banana")
count["a"]              # 3
count["x"]              # 0，不抛 KeyError
count.most_common(2)    # [('a', 3), ('n', 2)]

from collections import defaultdict

graph = defaultdict(list)
graph[1].append(2)
graph[1].append(3)

frequency = defaultdict(int)
frequency["apple"] += 1

items = [("apple", 3), ("banana", 1), ("pear", 3)]

# 数量降序，数量相同时名称升序
items.sort(key=lambda item: (-item[1], item[0]))

ordered = sorted(frequency.items(), key=lambda item: (-item[1], item[0]))