Python-algorithm

nums = [1, 2, 3, 4] 

ans = []

first = nums[0] # 正向索引，从0开始

last = nums[-1] # 反向索引，从-1开始

copy = nums[:] # 创建列表浅拷贝

reverse = nums[::-1] # 翻转列表

len(nums)          # 获取长度
sum(nums)          # 对数字列表求和
min(nums) / max(nums) # 求最值
sorted(nums)       # 返回一个排好序的新列表
value in nums      # 成员检查 (O(n))

nums.append(value)   # 在末尾添加
nums.pop(index)      # 删除并返回元素
nums.sort()          # 原地排序
nums.reverse()       # 原地反转
nums.index(value)    # 查找索引

lookup = {'name': 'Alice', 'id': 123}
empty_dict = {}

name = lookup['name'] # 不存在会报错
lookup['id'] = 456 # 添加新的键值对

'id' in lookup # 检查键是否存在

del lookup['id'] # 删除键值对

for key in lookup.keys() # 遍历所有键

for value in lookup.values() # 遍历所有值

for key, value in lookup.items() # 同时遍历键和值

unique_nums = {1, 2, 3}
empty_set = set() # 不可以使用{}

from_list = set([1, 2, 2, 3, 1]) # -> {1, 2, 3} (天然去重)

unique_nums.add(4) # 添加元素
unique_nums.remove(3) # 删除元素

3 in unique_nums # 检查成员是否存在

set1 = {1, 2, 3}
set2 = {3, 4, 5}

# 交集 (intersection) - 两个集合中都有的元素
intersection = set1 & set2  # -> {3}

# 并集 (union) - 两个集合中所有的元素
union = set1 | set2         # -> {1, 2, 3, 4, 5}

# 差集 (difference) - 在 set1 中但不在 set2 中的元素
difference = set1 - set2    # -> {1, 2}

new_str = str1 + str2

' '.join(['111', '222', '333']) # -> "111 222 333" (将列表连接成字符串)

"a,b,c".split(',') # -> ['a', 'b', 'c'] (将字符串分割成列表)

if score > 90:
    grade = 'A'
elif score > 80:
    grade = 'B'
else:
    grade = 'C'

for num in numbers:
    print(num)

# 使用 range() 进行固定次数的循环
for i in range(5): # 循环 0, 1, 2, 3, 4
    print(i)

# 使用 enumerate
for index, value in enumerate(numbers):
    print(f"Index: {index}, Value: {value}")

count = 5
while count > 0:
    print(count)
    count -= 1

def solve(parameter1, parameter2):
    # 1. 初始化变量
    result = 0
    
    # 2. 核心逻辑
    # ... (使用循环、判断等)
    
    # 3. 返回结果
    return result

length = len([1, 5, 9])      # -> 3
str_len = len("hello")       # -> 5
dict_len = len({'a': 1, 'b': 2}) # -> 2

# 适用于数字组成的列表、元组等
total = sum([10, 20, 30])    # -> 60

# 适用于可比较元素组成的序列
min_val = min([3, 1, 9, 2])  # -> 1
max_val = max([3, 1, 9, 2])  # -> 9
min_char = min("database")   # -> 'a' (按字母序)

# 返回一个全新的排好序的列表，不改变原对象
nums = [3, 1, 4, 2]
new_sorted_list = sorted(nums) # -> [1, 2, 3, 4]
# print(nums) 仍然是 [3, 1, 4, 2]

# 对字符串排序会得到字符列表
sorted_chars = sorted("bca") # -> ['a', 'b', 'c']

num = abs(-5) # 5

# range(stop)
for i in range(3):
    print(i) # -> 依次输出 0, 1, 2

# range(start, stop)
for i in range(1, 4):
    print(i) # -> 依次输出 1, 2, 3

# range(start, stop, step)
for i in range(0, 5, 2):
    print(i) # -> 依次输出 0, 2, 4

a=int("123") # -> 123 

a=str(123) # -> "123"

ans=list(range(3)) # -> [0, 1, 2]

new_set=set([1, 2, 2, 3]) # -> {1, 2, 3}

# 在循环中同时需要下标和元素的最佳方式
letters = ['a', 'b', 'c']
for index, value in enumerate(letters):
    print(f"Index: {index}, Value: {value}")
# -> Index: 0, Value: a
# -> Index: 1, Value: b
# -> Index: 2, Value: c

# 将多个列表/元组等并行打包遍历
names = ['Alice', 'Bob']
scores = [95, 88]
for name, score in zip(names, scores):
    print(f"{name}: {score}")
# -> Alice: 95
# -> Bob: 88 

# 将函数应用于序列的每个元素，返回一个迭代器
str_nums = ["1", "2", "3"]
# 需要用 list() 来获取所有结果
int_nums = list(map(int, str_nums)) # -> [1, 2, 3]


line = "10 20 30"
nums = list(map(int, line.split())) # -> [10, 20, 30]

s = "hello world"
words = s.split(' ')     # -> ['hello', 'world']

csv = "a,b,c"
items = csv.split(',')   # -> ['a', 'b', 'c']

words = ['hello', 'world']
s = " ".join(words)      # -> "hello world"

chars = ['p', 'y']
result = "-".join(chars) # -> "p-y"

s = "banana"
# 找不到时返回 -1
index1 = s.find('na')    # -> 2 (首次出现的位置)
index2 = s.find('z')     # -> -1

s = "banana"
count = s.count('a')     # -> 3

s = "  hello  "
clean_s = s.strip()      # -> "hello"

s1 = "123"
s2 = "a123"
print(s1.isdigit())      # -> True
print(s2.isdigit())      # -> False

aList = [123, 'xyz', 'runoob', 'abc']  
  
index1 = aList.index( 'xyz' )   # 1
index2 = aList.index( 'runoob', 1, 3 ) # 2

nums = [1, 2]
nums.append(3)           # nums 变为 [1, 2, 3]

nums = [10, 20, 30]
last = nums.pop()        # -> 30, nums 变为 [10, 20]
# 删除并返回指定索引的元素
first = nums.pop(0)      # -> 10, nums 变为 [20]

nums = [3, 1, 4, 2]
nums.sort()              # nums 本身变为 [1, 2, 3, 4]

# 降序排序
nums.sort(reverse=True)  # nums 变为 [4, 3, 2, 1]

# 获取列表的第二个元素
def takeSecond(elem):
    return elem[1]
random = [(2, 2), (3, 4), (4, 1), (1, 3)]
# 指定第二个元素排序
random.sort(key=takeSecond) # 排序列表： [(4, 1), (2, 2), (1, 3), (3, 4)]

nums = [1, 2, 3]
nums.reverse()           # nums 变为 [3, 2, 1]

list1 = ['Google', 'Runoob', 'Taobao', 'Baidu']  
list1.remove('Taobao')  # ['Google', 'Runoob', 'Baidu']
list1.remove('Baidu')   # ['Google', 'Runoob']

counts = {'a': 2, 'b': 1}
# 获取键 'b' 的值
val1 = counts.get('b', 0)  # -> 1
# 获取键 'c' 的值，因不存在，返回默认值 0
val2 = counts.get('c', 0)  # -> 0

counts = {'a': 2, 'b': 1}
for key in counts.keys():
    print(key) # -> 依次输出 'a', 'b'

sorted_keys = sorted(counts.keys()) # -> ['a', 'b']

counts = {'a': 2, 'b': 1}
for value in counts.values():
    print(value) # -> 依次输出 2, 1

values_list = list(counts.values()) # -> [2, 1]

counts = {'a': 2, 'b': 1}
for key, value in counts.items():
    print(f"{key}: {value}")
# -> a: 2
# -> b: 1

dd = defaultdict(lambda: 'N/A')