【MIT 公开课】Computer Science and Programing Lession 8

第八节课主要讲的是算法的复杂度问题：

一.求a^b:

1.

def exp1(a,b):
ans = 1
while (b>0):
ans *= a
b -= 1
return ans

T(b)=3b+2

O(b) 线性复杂度

2.

def exp2(a,b):
if b == 1:
return a
else: return a*exp2(a,b-1)

递归法：

T(b)=3+T(b-1)

      =3*k+T(b-k)  当b-k=1时停止

      =3*（b-1）+2

      =3b-1

[b]O(b) 线性复杂度[/b]

   3.

def exp3(a,b):
if b == 1:
return a
if (b%2)*2 == b:
return exp3(a*a, b/2)
else: return a*exp3(a,b-1)

   当b为偶数时：T(b)=6+T(b/2)

   当b为奇数时：T(b)=6+T(b-1)=12+T[(b-1)/2]

所以：T(b)=12+12+T(b/2^2)=12*k+T(b/2^k)

当2^k=b时停止，所以k=log2^b

O(logb)对数级复杂度

二.

def g(n):
x = 0
for i in range(n):
for j in range(m):
x += 1
return x

内嵌循环的复杂度

T=n*m

当m=n时，复杂度为O（n^2）

平方级的复杂度

三.

def Towers(size,fromStack,toStack,spareStack):
if size == 1:
print 'Move disk from ',fromStack, 'to ',toStack
else:
Towers(size-1,fromStack,spareStack,toStack)
Towers(1,fromStack,toStack,spareStack)
Towers(size-1,spareStack,toStack,fromStack)

汉诺塔问题


T(n)=1+2T(n-1)+T(1)=3+2T(n-1)=3+2*3+4T(n-2)=3(1+2+4+……+2^(k-1))+2^kT(n-k)

复杂度为O(Z^n)

指数级复杂度

四.从排好序的数组中查找某个元素

1.

def search(s, e):
answer = None
i = 0
numCompares = 0
while i < len(s) and answer == None:
numCompares += 1
if e == s[i]:
answer = True
elif e < s[i]:
answer = False
i += 1
print answer, numCompares

遍历法：O（len(s)）线性级复杂度

2.def bsearch(s, e, first, last):
print first, last
if (last - first) < 2: return s[first] == e or s[last] == e
mid = first + (last - first)/2
if s[mid] == e: return True
if s[mid] > e: return bsearch(s, e, first, mid - 1)
return bsearch(s, e, mid + 1, last)二分法：复杂度为对数级，每次把规模减小一半。