python簡單演算法整理

氣泡排序

def func(alist):

for x in range(1,len(alist)):
for i in range(0,len(alist)-x):
if alist[i] > alist[i+1]:
alist[i], alist[i+1] = alist[i+1], alist[i]
return alist
print(func([1,4,2,3,6,7,8,9,0,5]))

def func(alist):

for x in range(0,len(alist)):
min_num = alist[x]
for i in range(x+1,len(alist)):
if alist[i] < min_num:
alist[i], min_num = min_num, alist[i]
alist[x] = min_num
return alist
print(func([1,4,2,3,6,7,8,9,0,5]))

def func(n):

a = 0
b = 1
alist = 
if n <= 2:
return alist
else:
for i in range(n):
a, b = b, a + b
return alist
print(func(9))

def func(alist, item):

low = 0
high = len(alist)-1
n = 0
while low <= high:
mid = int((low + high)/2)
n += 1
if alist[mid]==item:
return mid
if alist[mid]鍊錶
class node:

def __init__(self, initdata):
self.__data = initdata
self.__next = none
def getdata(self):
return self.__data
def getnext(self):
return self.__next
def setdata(self, newdata):
self.__data = newdata
def setnext(self, newnext):
self.__next = newnext
class sincyclinkedlist:
def __init__(self):
self.head = node(none)
self.head.setnext(self.head)
def add(self, item):
temp = node(item)
temp.setnext(self.head.getnext())
self.head.setnext(temp)
def remove(self, item):
prev = self.head
while prev.getnext() != self.head:
cur = prev.getnext()
if cur.getdata() == item:
prev.setnext(cur.getnext())
prev = prev.getnext()
def search(self, item):
cur = self.head.getnext()
while cur != self.head:
if cur.getdata() == item:
return true
cur = cur.getnext()
return false
def empty(self):
return self.head.getnext() == self.head
def size(self):
count = 0
cur = self.head.getnext()
while cur != self.head:
count += 1
cur = cur.getnext()
return count
if __name__ == '__main__':
s = sincyclinkedlist()
print('s.empty() == %s, s.size() == %s' % (s.empty(), s.size()))
s.add(19)
s.add(86)
print('s.empty() == %s, s.size() == %s' % (s.empty(), s.size()))
print('86 is%s in s' % ('' if s.search(86) else ' not',))
print('4 is%s in s' % ('' if s.search(4) else ' not',))
print('s.empty() == %s, s.size() == %s' % (s.empty(), s.size()))
s.remove(19)
print('s.empty() == %s, s.size() == %s' % (s.empty(), s.size()))

乙隻青蛙一次可以跳上1級台階，也可以跳上2級……它也可以跳上n級。求該青蛙跳上乙個n級的台階總共有多少種跳法。

fib = lambda n: n if n < 2 else 2 * fib(n - 1)

#! /usr/bin/env python

# -*- coding: utf-8 -*-
def quick(list):
if len(list) < 2:
return list
tmp = list[0] # 臨時變數 可以取隨機值
left = [x for x in list[1:] if x <= tmp] # 左列表
right = [x for x in list[1:] if x > tmp] # 右列表
return quick(left) + [tmp] + quick(right)
li = [4,3,7,5,8,2]
print quick(li) # [2, 3, 4, 5, 7, 8]
#### 對[4,3,7,5,8,2]排序
'''[3, 2] + [4] + [7, 5, 8] # tmp = [4]
[2] + [3] + [4] + [7, 5, 8] # tmp = [3] 此時對[3, 2]這個列表進行排序
[2] + [3] + [4] + [5] + [7] + [8] # tmp = [7] 此時對[7, 5, 8]這個列表進行排序
'''

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