Python 百题大冲关

训练营

3537 人学过 392 次评价作者: xnmzdxws 难度: 中级综合评分: 8.7

课程包含 100 个 Python 编程练习题目，覆盖了基本语法，数据结构，算法等多个方面，是你熟悉 Python 编程和面试刷题的必备宝典。所有用户均可免费试学前 20 个挑战，高级会员可免费学习完整内容。

你将学到的

数组和字符串
堆栈和队列
排序算法
数学与概率

链表
图和树
递归和动态编程
位操作

课程介绍
课程内容

课程包含 100 个 Python 编程练习题目，覆盖了基本语法，数据结构，算法等多个方面，是你熟悉 Python 编程和面试刷题的必备宝典。所有用户均可免费试学前 20 个挑战，高级会员可免费学习。欢迎大家通过实验报告分享你自己的挑战解法。

图片描述

参考答案

参考答案仓库：https://github.com/lanqiao-courses/python-100

版权声明

课程改编自 Donne Martin 的开源项目，该项目使用 Apache 2.0 LICENSE。在原 LICENSE 的许可范围内，我们对原英文题目内容进行了编译以便于理解，同时修改了部分解题和单元测试代码以适配蓝桥云课在线环境。

课程信息

学习周期：课程有效期 2 个月，2 个月后无法进行在线实验，但可查看文档内容。高级会员有效期内可一直学习，到期后课程失效，且无法查看文档。部分云主机实验环境因成本较高，存在开启次数限制。关于课程退款等相关注意事项说明，请阅读用户付费协议。
版权说明：课程内容为蓝桥云课原创或蓝桥云课在原作者授权下制作。未经书面同意，擅自爬取、转载和再分发课程内容，均将受到严肃追责。
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最新评论

综合评分:

8.7

LOU1502373525

L2

忘了好多Python的关键字...

2021-09-25 21:39:49 来自：确定字符串是否包含唯一字符

考鸡蛋

L216

利用set去重的特性，通过比较去重后的长度和实际长度来判断string中是否包含重复的字符

2021-09-22 15:36:49 来自：确定字符串是否包含唯一字符

小优

L6

class UniqueChars(object):

def has_unique_chars(self, string):
    if string==None: return False
    else: return len(set(string)) == len(string)

2021-08-30 14:40:51 来自：确定字符串是否包含唯一字符

G.Sir

L133

from linked_list import LinkedList

class MyLinkedList(LinkedList):

def is_palindrome(self):
    ### 补充代码 ###
    origin_data = []
    reverse_data = []
    if self.head is None or self.head.next is None:
        return False

    cur = self.head
    while cur != None:
        origin_data.append(cur.data)
        reverse_data.insert(0, cur.data)
        cur = cur.next
    return origin_data == reverse_data

2021-08-24 17:01:57 来自：确定链表值是否为回文结构

G.Sir

L133

from linked_list import LinkedList

class MyLinkedList(LinkedList):

def kth_to_last_elem(self, k):
    length = len(self)

    if self.head is None or k >= length:
        return None

    k_node = length - k 

    cur = self.head
    count = 1
    while cur and count != k_node:
        cur = cur.next
        count += 1
    return cur.data

m = MyLinkedList() m.append(1) m.append(1) m.append(12) m.append(1) m.append(1) m.append(12) print(m.kth_to_last_elem(3))

2021-08-24 15:36:19 来自：寻找链表倒数第 k+1 个结点

G.Sir

L133

from linked_list import LinkedList

class MyLinkedList(LinkedList):

def remove_dupes(self):
    # cur = self.head
    # while cur and cur.next:
    #     if cur.data == cur.next.data:
    #         cur.next = cur.next.next
    #     else:
    #         cur = cur.next
    
    data = self.head
    cur = data
    while data:
        while cur and cur.next:
            if cur.next.data == data.data:
                cur.next = cur.next.next
            else:
                cur = cur.next
        data = data.next
        cur = data

l = MyLinkedList() l.append(1) l.append(2) l.append(1) l.append(2) l.append(2) l.append(3) print(l.get_all_data()) l.remove_dupes() print(l.get_all_data())

2021-08-24 15:28:55 来自：删除链表的重复项

G.Sir

L133

class Node(object): def init(self, data, next_node=None): self.data = data self.next_node = next_node def str(self): return self.data class LinkedList(object): def init(self, head=None): self.head = head def len(self): cur = self.head count = 0 while cur != None: count += 1 cur = cur.next_node return count

def insert_to_front(self, data):
    if data is None:
        return None
    node = Node(data)
    node.next_node = self.head
    self.head = node
    return node

def append(self, data):
    if data is None:
        return None

    node = Node(data)
    if self.head is None:
        self.head = node
        return node
    else:
        cur = self.head
        while cur.next_node != None:
            cur = cur.next_node
        cur.next_node = node
        return node

def find(self, data):
    cur = self.head
    while cur != None:
        if cur.data == data:
            return data
        cur = cur.next_node
    return None

def delete(self, data):
    cur = self.head
    if cur and cur.data == data:
        self.head = self.head.next_node
    else:
        while cur != None and cur.next_node != None:
            if cur.next_node.data == data:
                cur.next_node = cur.next_node.next_node
            cur = cur.next_node
def print_list(self):
    cur = self.head
    while cur != None:
        print(cur.data)
        cur = cur.next_node

def get_all_data(self):
    l = []
    cur = self.head
    while cur != None:
        l.append(cur.data)
        cur = cur.next_node
    return l

2021-08-24 15:03:55 来自：实现链表类

G.Sir

L133

class Solution(object): def fizz_buzz(self, num): if num is None: raise TypeError("") if num < 1: raise ValueError("")

    result = []
    for i in range(1, num+1):
        if i % 3 == 0 and i % 5 == 0:
            result.append("FizzBuzz")
        elif i % 3 == 0:
            result.append("Fizz")
        elif i % 5 == 0:
            result.append("Buzz")
        else:
            result.append(str(i))

    return result

s = Solution().fizz_buzz(4) print(s)

2021-08-23 23:30:48 来自：Fizz Buzz 经典问题

G.Sir

L133

class Solution(object): def find_diff(self, str1, str2): if str1 is None or str2 is None: raise TypeError("")

    if len(str1) < len(str2):
        str1, str2 = str2, str1

    # 保证str1是长的

    for i in str1:
        if str1.count(i) != str2.count(i):
            return i

2021-08-23 22:45:08 来自：找到给定字符串中的不同字符

G.Sir

L133

题目都描述不清楚，样例都没有，？。

class ReverseString(object): def reverse(self, chars): if chars is None or not chars: return None for i in range(len(chars)//2): print(i) print(len(chars)-i-1) chars[i], chars[len(chars)-i-1] = chars[len(chars)-1-i], chars[i] return chars print(ReverseString().reverse( ['f', 'o', 'o', ' ', 'b', 'a', 'r']))

2021-08-23 22:41:32 来自：反转字符串中的字符

G.Sir

L133

class CompressString(object):

def compress(self, string):
    if string is None or not string:
        return string
    
    s = ""
    pre = string[0]
    count = 1

    for i in string[1:]:
        # print(i, end=", ")
        if pre == i:
            count += 1
        else:
            if count > 1:
                s += pre + str(count)
            else:
                s += pre
            pre = i
            count = 1
    else:
        if count > 1:
            s += pre + str(count)
        else:
            s += pre

    # print(s)
    return s if len(string) > len(s) else string

print(CompressString().compress(None) == None) print(CompressString().compress('') == '') print(CompressString().compress('AABBCC') == 'AABBCC') print(CompressString().compress('AAABCCDDDDE') == 'A3BC2D4E') print(CompressString().compress('BAAACCDDDD') == 'BA3C2D4') print(CompressString().compress('AAABAACCDDDD') == 'A3BA2C2D4')

2021-08-23 22:05:11 来自：压缩字符串

G.Sir

L133

``` class Rotation(object): def is_sbustring(self, s1, s2): return s1 in s2

def is_rotation(self, s1, s2):
    if s1 is None or s2 is None:
        return False

    if len(s1) != len(s2):
        return False

    return self.is_sbustring(s1, s2+s2)

看不懂的题目操作

2021-08-23 19:57:19 来自：确定字符串是否是另一个的旋转

G.Sir

L133

class Permutations(object): def is_permutation(self, str1, str2): if str1 is None or str2 is None: return False

    return sorted(str1) == sorted(str2)

2021-08-23 19:49:04 来自：确定字符串是否是另一个的排列

G.Sir

L133

class UniqueChars(object): def has_unique_chars(self, string): if string is None: return False for i in string: if string.count(i) > 1: return False else: return True

2021-08-23 19:46:10 来自：确定字符串是否包含唯一字符

玖常科技

L19

应用字典： dictnums = {} for i, x in enumerate(nums): if val-x in dictnums: return [dictnums[val-x], i] else: dictnums[x] = i return None

2021-06-02 12:56:28 来自：查找两个总和为特定值的索引