#!/usr/bin/python # -*- coding: UTF-8 -*- # 例2：elif用法 num = 5 if num == 3 : # 判断num的值 print ' boss ' elif num == 2 : print ' user ' elif num == 1 : print ' worker ' elif num < 0 : # 值小于零时输出 print ' error ' else : print ' roadman ' # 条件均不成立时输出

#!/usr/bin/python # -*- coding: UTF-8 -*- # 例3：if语句多个条件 num = 9 if num >= 0 and num <= 10 : # 判断值是否在0~10之间 print ' hello ' # 输出结果: hello num = 10 if num < 0 or num > 10 : # 判断值是否在小于0或大于10 print ' hello ' else : print ' undefine ' # 输出结果: undefine num = 8 # 判断值是否在0~5或者10~15之间 if ( num >= 0 and num <= 5 ) or ( num >= 10 and num <= 15 ) : print ' hello ' else : print ' undefine ' # 输出结果: undefine

#!/usr/bin/python # -*- coding: UTF-8 -*- var = 100 if ( var == 100 ) : print " 变量 var 的值为100 " print " Good bye! "

#!/usr/bin/env python
# -*- coding: UTF-8 -*-
# 一个简单的条件循环语句实现汉诺塔问题
def my_print(args):
    print args
def move(n, a, b, c):
    my_print ((a, '-->', c)) if n==1 else (move(n-1,a,c,b) or move(1,a,b,c) or move(n-1,b,a,c))
move (3, 'a', 'b', 'c')

#!/usr/bin/env python
# -*- coding: utf-8 -*-
import os
def zero():
    return "zero"
def one():
    return "one"
def two():
    return "two"
def num2Str(arg):
    switcher={
        0:zero,
        1:one,
        2:two,
        3:lambda:"three"
    func=switcher.get(arg,lambda:"nothing")
    return func()
if __name__ == '__main__':
    print num2Str(0)

#!/usr/bin/python
# -*- coding: UTF-8 -*-
a = [1,2,3]
b = a if len(a) != 0 else ""
print(b)
d = c if len(c) != 0 else "c 是一个空列表"
print(d)

[1, 2, 3]
c 是一个空列表

def deduplication(nums):
    # write your code here
    exist_nums = {} 
    pointer = 0
    for num in nums:
        if num not in exist_nums:
            exist_nums[num] = True
            nums[pointer] = num
            pointer += 1
    return pointer
print(deduplication([1,1,1,1,1,1,2,2,2,2,2,2,2,2]))