Python--實現二分法算法

• 專業分析二分法查找

•  二分法針對的對象要是有序的隊列,通過索引找到你要查找的值
•  思想:取一個索引開頭left,取一個索引末尾(len(隊列)-1)right,讓你的值跟(left + right)//2(middle)比
• 如果索引middle值比你查的值大,把middle賦值給right,反之把middle賦值給left;
• 代碼展示:

• lst = [11,22,33,44,55,66,77,88,99,123,234,345,456,567,678,789,1111]
  n = 456
  left = 0
  right = len(lst) - 1
  count = 1
  while left <= right:
      middle = (left + right) // 2
      if n > lst[middle]:
          left = middle + 1
      elif n < lst[middle]:
          right = middle - 1
      else:
          print('查找次數%s' % count)
          print("存在位置%s" % middle)
          break
      count = count + 1
  else:
      print("不存在")

  用遞歸方法實現二分法:

"""
其實二分法就是兩個點
    1: 函數開始前,要先判斷滿足條件, 左索引值要小於右索引值
    2: 就是比較之後, 左右索引要移動到中間值的前一位或者後一位索引位置上
""" 
def binary_search(left, right, n):
    middle = (left + right)//2
    if left > right:
        return -1
    if n > lst[middle]:
        left = middle + 1
    elif n < lst[middle]:
        right = middle - 1
    else:
        return middle
    return binary_search(left, right, n)
print(binary_search(0, len(lst)-1, 65) )

---

迭代器50000000000000000星級難題

------問題-------
def add(a, b):
    return a + b


def test():
    for i in range(4):
        yield i


g = test()
for n in [2,10]:
    g = (add(n,i) for i in g)
print(list(g))  #[20,21,22,23]

######## 拆分講解
def add(a, b):
    return a + b


def test():
    for i in range(4):
        yield i

g = test()

#-----循環體
for n in [2,10]:
    g = (add(n,i) for i in g)
print(list(g))

#-----對循環體拆分
# n=2時
n = 2
g = (add(n, i) for i in g) 
# n=10時
n = 10
g = (add(n, i) for i in g)


# -----所以for循環那一刻可以縮寫代碼如下
g = (add(n,i) for i in (add(n,i) for i in test()))
所以在一直沒有list(g)取值時,生成器只會先生成表達式,只有要值了,纔會把test()帶進去值

---

62進制的問題:

s1 = ''.join([chr(i) for i in range(65,90)])
s2 = ''.join([chr(i) for i in range(97,122)])
s3 = ''.join([str(i) for i in range(10)])
s = s3+s1+s2
res = ''
while n!=0:
    ord_ = n%62
    n = n//62
    res += s[ord_]
print(res)