python中如何計(jì)算列表中元素的個(gè)數(shù)？描述len() 方法返回列表元素個(gè)數(shù)。語(yǔ)法len()方法語(yǔ)法：len(list)參數(shù)list -- 要計(jì)算元素個(gè)數(shù)的列表。返回值返回列表元素個(gè)數(shù)。實(shí)例以下實(shí)例展

python中如何計(jì)算列表中元素的個(gè)數(shù)？

描述len() 方法返回列表元素個(gè)數(shù)。語(yǔ)法len()方法語(yǔ)法：len(list)參數(shù)list -- 要計(jì)算元素個(gè)數(shù)的列表。返回值返回列表元素個(gè)數(shù)。實(shí)例以下實(shí)例展示了 len()函數(shù)的使用方法：#!/usr/bin/pythonlist1， list2 = [123， "xyz"， "zara"]， [456， "abc"]print "First list length : "， len(list1)print "Second list length : "， len(list2)以上實(shí)例輸出結(jié)果如下：First list length : 3Second lsit length : 2

python text中按字典序排列最小的子序列？

class Solution(object):

def smallestSubsequence(self， text):

"""

:type text: str

:rtype: str

"""

stack = []

last_o = {}

considered = {}

for i in range(len(text)-1，-1，-1):

if text[i] not in last_o:

last_o[text[i]] = i

considered[text[i]] = False

print(last_o)

i = 0

while i < len(text):

print(stack，i，text[i])

if len(stack) == 0:

stack.append(text[i])

considered[text[i]] = True

i =1

elif stack[-1]>text[i] and considered[text[i]] == False:

if last_o[stack[-1]]>i:

considered[stack[-1]]=False

stack.pop()

else:

considered[text[i]] = True

stack.append(text[i])

i =1

elif stack[-1]<text[i] and considered[text[i]] == False:

stack.append(text[i])

considered[text[i]] = True

i =1

else:

i =1

return "".join(i for i in stack)