【力扣hot100】刷题笔记Day1

前言

• 既然打算年后去找算法的实习，所以之后想直接改用python刷hot100了（新坑芜湖~），在B站大学找到这个刷题教程，先快速过一遍里面提到的python语法

Python数组

• # 创建数组
  a = []
  # 添加元素，O(1)
  a.append(1)
  a.append(2)
  a.append(3)  # [1, 2, 3]
  # 插入元素，O(N)
  a.insert(2,99)  # [1, 2, 99, 3]
  # 访问元素，O(1)
  temp = a[2]  # 99
  # 更新元素，O(1)
  a[2] = 88  # 88
  # 删除元素，O(N)
  a.remove(88)  # [1, 2, 3]
  a.pop(1)  # 索引值，O(N)  # [1, 3]
  a.pop()  # 尾部，O(1)  # [1]
  # 获取数组长度
  size = len(a)
  # 遍历数组，O(N)
  for i in a:
      print(i)
  for index, element in enumerate(a):
      print("Index at ", index, "is ", element)
  for i in range(0, len(a)):
      print("i: ", i, "element: ", a[i])
  # 查找元素，O(N)
  index = a.index(2)  # 查找元素2所对应的索引
  # 数组排序，O(NlogN)
  a = [3, 1, 2]
  a.sort()  # a = [1, 2, 3]
  a.sort(reverse=True)  # a = [3, 2, 1]

Python链表

• # 创建链表，O(1)
  linkedlist = deque()
  # 添加元素
  linkedlist.append(1)    # O(1)
  linkedlist.append(2)
  linkedlist.append(3)    # [1, 2, 3]
  linkedlist.append(2, 99) # O(N)  # [1, 2, 99, 3]
  # 访问元素，O(1)
  element = linkedlist[2]  # 99
  # 搜索元素，O(N)
  index = linkedlist.index(99)  # 查找元素99所对应的索引，2
  # 更新元素，O(N)
  linkedlist[2] = 88  # [1, 2, 88, 3]
  # 删除元素，O(N)
  del linkedlist[2]  # 删除索引为2的元素
  linkedlist.remove(88)  # 删除值为88的元素
  # 长度
  lenth = len(linkedlist)

Python队列

• # 创建队列
  queue = deque()
  # 添加元素，O(1)
  queue.append(1)
  queue.append(2)
  queue.append(3)  # [1, 2, 3]
  # 获取即将出队元素，O(1)
  temp1 = queue[0]  # 1
  # 删除即将出队的元素，O(1)
  temp2 = queue.popleft()  # 1, [2, 3]
  # 队列长度判空，O(1)
  len(queue) == 0
  # 边删除边遍历队列，O(N)
  while len(queue) != 0:
      temp = queue.popleft()

Python栈

• # 创建栈
  stack = []
  # 添加元素，O(1)
  stack.append(1)
  stack.append(2)
  stack.append(3)  # [1, 2, 3]
  # 获取栈顶元素，O(1)
  stack[-1]  # 3
  # 删除栈顶元素，O(1)
  temp = stack.pop()  # 2
  # 栈是否为空，O(1)
  len(stack) == 0
  # 边删除边遍历栈，O(N)
  while len(stack) > 0:
      temp = stack.pop()

Python哈希表

• # 数组创建哈希表
  hashTable = ['']*4
  # 字典创建哈希表
  mapping = {}
  # 添加元素，0(1)
  hashTable[1] = 'hanmeimei'
  hashTable[2] = 'lihua'
  hashTable[3] = 'siyangyuan'
  mapping[1] = 'hanmeimei'
  mapping[2] = 'lihua'
  mapping[3] = 'siyangyuan'
  # 修改元素，0(1)
  hashTable[1] = 'bishi'
  mapping[1] = 'bishi'
  # 删除元素，0(1)
  hashTable[1] = ''
  mapping.pop(1)
  del mapping[1]
  # 获取元素，0(1)
  hashTable[3]
  mapping[3]
  # 检查key是否存在，0(1)
  3 in mapping
  # 哈希表长度判空，0(1)
  len(mapping) == 0

 Python集合

• # 创建集合
  s = set()
  # 添加元素，O(1)
  s.add(10)
  s.add(3)
  s.add(5)
  s.add(2)  # 顺序不定，自动去重
  s.add(2)  # {2,10,3,5}
  # 搜索元素，O(1)
  2 in s
  # 删除元素，O(1)
  s.remove(2)  # {10,3,5}
  # 长度，O(1)
  len(s)

Python堆

• import heapq
  # 创建最小堆
  # 最大堆的话需要元素都取反，按最小堆算，结果再取反
  minheap = []
  heapq.heapify(minheap)
  # 增加元素
  heapq.heappush(minheap,10);
  heapq.heappush(minheap,8);
  heapq.heappush(minheap,9);
  heapq.heappush(minheap,2);
  heapq.heappush(minheap,1);
  heapq.heappush(minheap,11);  # [1,2,9,10,8,11]
  # 堆顶元素
  minheap[0]
  # 删除堆顶，自动调整
  heapq.heappop(minheap)
  # 大小
  len(minheap)
  # 迭代
  while len(minheap) != 0:
      print(heapq.heappop(minheap))

后言

•  下午又要干活+备课了，在家的时间利用要充实起来！