#Notes:#Both num1 and num2 contains only digits 0-9.#Both num1 and num2 does not contain any leading zero.
num1 = '364'num2 = '1836'
# Approach 1:def solution(num1,num2):eval(num1) + eval(num2)return str(eval(num1) + eval(num2))
print(solution(num1,num2))
#Approach2#Given a string of length one, the ord function returns#an integer representing the Unicode code point of the character#when the argument is a unicode object, or the value of#the byte when the argument is an 8-bit string.
def solution(num1, num2):n1, n2 = 0, 0m1, m2 = 10**(len(num1)-1), 10**(len(num2)-1)
for i in num1:n1 += (ord(i) - ord("0")) * m1m1 = m1//10
for i in num2:n2 += (ord(i) - ord("0")) * m2m2 = m2//10
return str(n1 + n2)print(solution(num1, num2))
Output:
2200
2200
【解析】
使用eval( )或者ord( )都能够解决上面的问题，前一个方法比较简单快捷，但我们还是建议使用后一种方法，能够从底层去深入了解字符串的ASCII及相应的字符格式与整数的相互转换，而且能够掌握进位计数制的构成方法。
4、非重复首字母
# Given a string, find the first non-repeating character in it# and return its index.# If it doesn't exist, return -1.# Note: all the input strings are already lowercase.#Approach 1def solution(s):frequency = {}for i in s:if i not in frequency:frequency[i] = 1else:frequency[i] +=1for i in range(len(s)):if frequency[s[i]] == 1:return ireturn -1
print(solution('alphabet'))print(solution('barbados'))print(solution('crunchy'))
print('###')
#Approach 2import collections
def solution(s):# build hash map : character and how often it appearscount = collections.Counter(s)# <-- gives back a dictionary with words occurrence count#Counter({'l': 1, 'e': 3, 't': 1, 'c': 1, 'o': 1, 'd': 1})# find the indexfor idx, ch in enumerate(s):if count[ch] == 1:return idxreturn -1
print(solution('alphabet'))print(solution('barbados'))print(solution('crunchy'))
Output:
1
2
1
###
1
2
1
【解析】
我们给出了2种解决的算法,如果你是Python的新手,建议你使用第1种方案(先使用一个空字典,然后给键加上一个计数功能值);长远来说,第2种方法,我们只是简单使用了collection.Counter(s)来替换字符计数,enumerate(s)替换range(len(s)),让程序列简短而优雅.
5、回文字符串验证
# Given a non-empty string s, you may delete at most one character. Judge whether you can make it a palindrome.# The string will only contain lowercase characters a-z.s = 'radkar'def solution(s):for i in range(len(s)):t = s[:i] + s[i+1:]if t == t[::-1]: return True
return s == s[::-1]
solution(s)
Output:
True
【解析】
回文是一个比较精典的题目,这个算法的核心是移除一个字符,让文本变成回文形式的文本,内容比较比较简单只要把文本进行反转即可.例如, s = ‘radkar’ 代入函数,当移除字符d时,得到的文本rakar就是回文格式,所以返回True.
6、递增或递减数列
# Given an array of integers, determine whether the array is monotonic or not.A = [6, 5, 4, 4]B = [1,1,1,3,3,4,3,2,4,2]C = [1,1,2,3,7]def solution(nums):return (all(nums[i] <= nums[i + 1] for i in range(len(nums) - 1)) orall(nums[i] >= nums[i + 1] for i in range(len(nums) - 1)))
print(solution(A))print(solution(B))print(solution(C))
Output:
True
False
True
【解析】
这个问题也是很多人会问及的,解决办法也是相当简单,我们只要使用单行就可以解决问题.数组中元素的单调性(递增或者递减).我们使用all函数来测试所有的枚举都成立(返回True),则函数返回True否则返回False.如果序列中没有元素,也返回True.
7、移动零元素
#Given an array nums, write a function to move all zeroes#to the end of it while maintaining the relative order of#the non-zero elements.array1 = [0,1,0,3,12]array2 = [1,7,0,0,8,0,10,12,0,4]
def solution(nums):for i in nums:if 0 in nums:nums.remove(0)nums.append(0)return nums
solution(array1)solution(array2)
Output:
[1, 3, 12, 0, 0]
[1, 7, 8, 10, 12, 4, 0, 0, 0, 0]
【解析】
当涉及到数组时,.remove和.append是一种好兄弟,在这个题目中,我们找到0元素并移除它,在这个序列的后面添加它,从而完成算法功能.
8、填充空元素
# Given an array containing None values fill in the None values with# most recent non None value in the arrayarray1 = [1,None,2,3,None,None,5,None]def solution(array):valid = 0res = []for i in nums:if i is not None:res.append(i)valid = ielse:res.append(valid)return res
solution(array1)
Output:
[1, 1, 2, 3, 3, 3, 5, 5]
【解析】
这个题目在编写代码时要注意None元素的正确使用.
9、文字匹配问题
#Given two sentences, return an array that has the words that#appear in one sentence and not#the other and an array with the words in common.sentence1 = 'We are really pleased to meet you in our city'sentence2 = 'The city was hit by a really heavy storm'
def solution(sentence1, sentence2):set1 = set(sentence1.split)set2 = set(sentence2.split)
return sorted(list(set1^set2)), sorted(list(set1&set2))# ^ A.symmetric_difference(B), & A.intersection(B)
print(solution(sentence1, sentence2))
Output:
(['The','We','a','are','by','heavy','hit','in','meet','our',
'pleased','storm','to','was','you'],
['city', 'really'])
【解析】
这个算法挺简单的,我们会使用常用的集合的方法,如set , intersection or &和 symmetric_differenceor ^.这些方法非常好用,如果您是第一次使用,务必做好记录工作.
10、打印素数序列
# Given k numbers which are less than n, return the set of# prime number among them# Note: The task is to write a program to print all Prime numbers# in an Interval.# Definition: A prime number is a natural number greater than 1# that has no positive divisors other than 1 and itself.n = 35def solution(n):prime_nums = []for num in range(n):if num > 1: # all prime numbers are greater than 1for i in range(2, num):if (num % i) == 0:# if the modulus == 0 is means that the number can be divided by a number preceding itbreakelse:prime_nums.append(num)return prime_numssolution(n)
Output:
[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31]
【解析】
最后一个题目是比较经典的问题,大家只要熟悉 素数的定义,结合 Range函数及 求余操作,就会快速的推断出算法.返回搜狐，查看更多