# 序列求和
print(sum([1,2,3,4,5,6,7,8,9,]))
# sum函数的原型是sum(iterable,start).可以设置一个初始值
print(sum([1,2,3,4,5,6,7,8,9,],5))
# 展开2层的嵌套列表
iterable = [[1,2],[3,4],[5,6],[7,8]]
print(sum(iterable,[0]))# 后面的start可以为空
# sum函数与下列函数等效
start = [0]
for element in iterable:
    start += element
print(start)
start = [0]
# sum 函数可以看作reduce()的特殊情况
from functools import reduce
from operator import add
print(reduce(add,iterable,start))

import itertools
class Solution:
    """
    @param: nums: A list of integers.
    @return: A list of permutations.
    """
    def permute(self, nums):
        pailie = list(itertools.permutations(nums))  # 要list一下，不然它只是一个对象

        return pailie



so = Solution()
print(so.permute([1,2,3]))

class Solution(object):

    # @param nestedList a list, each element in the list
    # can be a list or integer, for example [1,2,[1,2]]
    # @return {int[]} a list of integer
    def flatten(self, nestedList):
        a = []
        for i in range(len(nestedList)):#用于判断给定列表中的元素是否含有列表（注：这一目的主要是为了结束递归的死循环）
            if type(nestedList[i]) == list:#如果有就将列表中的列表分别取出来放到新的列表中
                for j in range(len(nestedList)):
                    if type(nestedList[j]) != list:
                        a.append(nestedList[j])
                    else:
                        for k in range(len(nestedList[j])):
                            a.append((nestedList[j])[k])
                return Solution.flatten(self,a)#使用递归完成多次“套娃”的情况

        return nestedList#如果没有就直接返回这个列表



nums = [4,[3,[2,[1]]]]
so = Solution()
print(so.flatten(nums))

"""
输出结果：
[4, 3, 2, 1]
"""

class Solution:
    """
    @param nums: The integer array.
    @param target: Target to find.
    @return: The first position of target. Position starts from 0.
    """
    def binarySearch(self, nums, target):
            low = 0
            high = len(nums) - 1
            while low <= high:
                mid = low + (high - low) // 2
                if  nums[mid] == target:
                    while(mid >= 0):#新增加的while和if用于解决有重复数字的情况，并保证每次都输出第一次出现的target的下标
                        if nums[mid] != target:
                            break

                        mid = mid - 1
                    if mid <= -1:
                        return 0
                    return mid + 1#多减了一次，所以输出时+1
                elif  target > nums[mid]:
                    low = mid + 1
                else:
                    high = mid - 1

            return -1



nums = [4,5,9,12,13,14,15,15,18]
so = Solution()
print(so.binarySearch(nums,9))

j = 0
for i in itertools.product("12345","ABCDE"):
    print(i,end='')
    j = j+1


print("")
print("一共有{}种组合".format(j))

"""
输出结果为:
('1', 'A')('1', 'B')('1', 'C')('1', 'D')('1', 'E')('2', 'A')('2', 'B')('2', 'C')('2', 'D')('2', 'E')('3', 'A')('3', 'B')('3', 'C')('3', 'D')('3', 'E')('4', 'A')('4', 'B')('4', 'C')('4', 'D')('4', 'E')('5', 'A')('5', 'B')('5', 'C')('5', 'D')('5', 'E')
一共有25种组合

"""

nums = [1,2,3,4]
print(list(itertools.permutations(nums)))

"""输出结果：
[(1, 2, 3, 4), (1, 2, 4, 3), (1, 3, 2, 4), (1, 3, 4, 2), (1, 4, 2, 3), (1, 4, 3, 2), (2, 1, 3, 4), (2, 1, 4, 3), (2, 3, 1, 4), (2, 3, 4, 1), (2, 4, 1, 3), (2, 4, 3, 1), (3, 1, 2, 4), (3, 1, 4, 2), (3, 2, 1, 4), (3, 2, 4, 1), (3, 4, 1, 2), (3, 4, 2, 1), (4, 1, 2, 3), (4, 1, 3, 2), (4, 2, 1, 3), (4, 2, 3, 1), (4, 3, 1, 2), (4, 3, 2, 1)]
"""

nums = [1,2,3,4]
print(list(itertools.combinations(nums,2)))
"""
输出结果：

[(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)]
"""

a = "fadsfga"
print(sorted(a))

"""输出结果：
['a', 'a', 'd', 'f', 'f', 'g', 's']


"""

a = 'fnsadfgdsgsadfdsa'
b = a.count('f')
print(b)
"""
输出结果：
3
"""

## ord()函数# ord() 函数以一个字符作为参数，返回参数对应的 ASCII 数值，便于和后面捕捉的键位关联

print(ord('a'))
print(ord('A'))

"""
输出结果：
97
65
"""

## 二元表达式

init = 1

number = 100 if init else 0

print(number)

##三元表达
wide= -1

new_w = 299 if  wide>0 else 'sdf' if wide==0 else 28
print(new_w)
"""
输出结果

100
28
"""

## randrange---给定一个范围，获取范围之内的随机数
import random
print(random.randrange(100))

'''enumerate() 函数用于将一个可遍历的数据对象(如列表、元组或字符串)组合为一个索引序列，同时列出数据和数据下标，
'''
names = 'chenqiaochu'
for key, value in enumerate(names):
  print(key,value)
  
 '''output:
 '''

a = {'name' : '重庆','mather' : 'China' }
for key in a.keys():
    print(key, a[key])
'''输出结果：
name 重庆
mather China
'''

f = abs
print(f(-100))
"""
输出结果:
100
"""

abs =10
print(abs)
print(abs(-200))

"""
输出结果：
10
Traceback (most recent call last):
  File "/Users/bowenkei/Desktop/Python/项目2-数据可视化/test.py", line 3, in <module>
    print(abs(-200))
TypeError: 'int' object is not callable

"""
# 有结果可以看出，此时abs已经失去了原有函数的意思了

# 最简单的高阶函数
# 一个最简单的高阶函数
def add(x,y,f):
    return (f(x)+f(y))


print(add(500,20,abs))

"""
输出结果：
520

"""

class Solution:
    """
    @param A: A string
    @param B: A string
    @return: if string A contains all of the characters in B return true else return false
    """
    def compareStrings(self, A, B):
        # write your code here
        #此题采用了反向思维的方法，通过举出返回Flase的几种情况后，其余都是True
        
        if len(A) < len(B):
            return False
        if len(A) == len(B):
            return(sorted(A) == sorted(B))
        for i in B:
            if A.count(i) < B.count(i):
                return False
        return True

class Solution:
    """
    @param source:
    @param target:
    @return: return the index
    """
    def strStr(self, source, target):
        geshu=0
        a = len(source)
        b = len(target)
        if source =="" and target =="":#排除特殊情况
            return 0
        elif source !="" and target =="":#继续排除特殊情况
            return 0
        elif target not in source:#被查找字符串不在已知字符串的情况
            return -1
        else:#需要返回第一个出现位置的情况
            for i in range(a):
                if target[0] == source[i]:
                    for j in range(b):
                        if (i+j) < a:#避免字符串索引溢出
                            if source[i+j] == target[j]:
                                geshu = geshu + 1

                        if geshu == b:
                             return i

so = Solution()
print(so.strStr("tartarget","target"))