Python练习题集合+剑指offer

#coding=gbk
'''
Created on 2018年8月27日

@author: Somebody
'''
# 1. 输入某年某月某日，判断这是这一年的第几天
#特殊情况：闰年，如果输入日期大于2月，闰年需要加1 
year = int(input('year:\n')) month = int(input('month:\n')) day = int(input('day:\n')) #判断输入的天数是否正确 # 1, 3, 5, 7, 8, 10 腊 month_count = [31, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31] if ((year % 400 ==0) or (year % 4 ==0 and year % 100 != 0)): month_count.insert(1, 29) else: month_count.insert(1, 28) print(month_count) if(day > month_count[month - 1]): print('输入的天数是错误的') else: if (0< month <= 12): sum =0 for i in range(month -1): sum += month_count[i] else: print('data error') sum += day print('该天是该年的第%d 天\n' %sum) # year: # 2000 # month: # 3 # day: # 1 # [31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31] # 该天是该年的第61 天

2.打印出 5 种不同形式的九九乘法表

（输出的数字的形式类似于第 7 道题中的打印出图形的例子，这里可以将数字看做是 7 中的字符 * ）

#完整格式输出九九乘法表
for i in range(1, 10): for j in range(1, 10): result = i * j print('%d*%d = %2d' %(i, j, result), end = " ") print() # 1*1 = 1 1*2 = 2 1*3 = 3 1*4 = 4 1*5 = 5 1*6 = 6 1*7 = 7 1*8 = 8 1*9 = 9 # 2*1 = 2 2*2 = 4 2*3 = 6 2*4 = 8 2*5 = 10 2*6 = 12 2*7 = 14 2*8 = 16 2*9 = 18 # 3*1 = 3 3*2 = 6 3*3 = 9 3*4 = 12 3*5 = 15 3*6 = 18 3*7 = 21 3*8 = 24 3*9 = 27 # 4*1 = 4 4*2 = 8 4*3 = 12 4*4 = 16 4*5 = 20 4*6 = 24 4*7 = 28 4*8 = 32 4*9 = 36 # 5*1 = 5 5*2 = 10 5*3 = 15 5*4 = 20 5*5 = 25 5*6 = 30 5*7 = 35 5*8 = 40 5*9 = 45 # 6*1 = 6 6*2 = 12 6*3 = 18 6*4 = 24 6*5 = 30 6*6 = 36 6*7 = 42 6*8 = 48 6*9 = 54 # 7*1 = 7 7*2 = 14 7*3 = 21 7*4 = 28 7*5 = 35 7*6 = 42 7*7 = 49 7*8 = 56 7*9 = 63 # 8*1 = 8 8*2 = 16 8*3 = 24 8*4 = 32 8*5 = 40 8*6 = 48 8*7 = 56 8*8 = 64 8*9 = 72 # 9*1 = 9 9*2 = 18 9*3 = 27 9*4 = 36 9*5 = 45 9*6 = 54 9*7 = 63 9*8 = 72 9*9 = 81 print('左上角打印出九九乘法表') for i in range(1, 10): for j in range(i, 10): result = i * j print('%d * %d = %2d'%(i, j, result), end= " ") print() # 左上角打印出九九乘法表 # 1 * 1 = 1 1 * 2 = 2 1 * 3 = 3 1 * 4 = 4 1 * 5 = 5 1 * 6 = 6 1 * 7 = 7 1 * 8 = 8 1 * 9 = 9 # 2 * 2 = 4 2 * 3 = 6 2 * 4 = 8 2 * 5 = 10 2 * 6 = 12 2 * 7 = 14 2 * 8 = 16 2 * 9 = 18 # 3 * 3 = 9 3 * 4 = 12 3 * 5 = 15 3 * 6 = 18 3 * 7 = 21 3 * 8 = 24 3 * 9 = 27 # 4 * 4 = 16 4 * 5 = 20 4 * 6 = 24 4 * 7 = 28 4 * 8 = 32 4 * 9 = 36 # 5 * 5 = 25 5 * 6 = 30 5 * 7 = 35 5 * 8 = 40 5 * 9 = 45 # 6 * 6 = 36 6 * 7 = 42 6 * 8 = 48 6 * 9 = 54 # 7 * 7 = 49 7 * 8 = 56 7 * 9 = 63 # 8 * 8 = 64 8 * 9 = 72 # 9 * 9 = 81 print('打印出右上角的九九乘法表') for i in range(1, 10): for k in range(1, i): print(end = " ") for j in range(i, 10): result = i * j print('%d * %d = %2d'%(i, j, result), end= " ") print() # 打印出右上角的九九乘法表 # 1 * 1 = 1 1 * 2 = 2 1 * 3 = 3 1 * 4 = 4 1 * 5 = 5 1 * 6 = 6 1 * 7 = 7 1 * 8 = 8 1 * 9 = 9 # 2 * 2 = 4 2 * 3 = 6 2 * 4 = 8 2 * 5 = 10 2 * 6 = 12 2 * 7 = 14 2 * 8 = 16 2 * 9 = 18 # 3 * 3 = 9 3 * 4 = 12 3 * 5 = 15 3 * 6 = 18 3 * 7 = 21 3 * 8 = 24 3 * 9 = 27 # 4 * 4 = 16 4 * 5 = 20 4 * 6 = 24 4 * 7 = 28 4 * 8 = 32 4 * 9 = 36 # 5 * 5 = 25 5 * 6 = 30 5 * 7 = 35 5 * 8 = 40 5 * 9 = 45 # 6 * 6 = 36 6 * 7 = 42 6 * 8 = 48 6 * 9 = 54 # 7 * 7 = 49 7 * 8 = 56 7 * 9 = 63 # 8 * 8 = 64 8 * 9 = 72 # 9 * 9 = 81 print('打印出左下角的九九乘法表') for i in range(1, 10): for j in range(1, i+1): result = i * j print('%d * %d = %2d'%(j, i, result), end= " ") print() # 打印出左下角的九九乘法表 # 1 * 1 = 1 # 1 * 2 = 2 2 * 2 = 4 # 1 * 3 = 3 2 * 3 = 6 3 * 3 = 9 # 1 * 4 = 4 2 * 4 = 8 3 * 4 = 12 4 * 4 = 16 # 1 * 5 = 5 2 * 5 = 10 3 * 5 = 15 4 * 5 = 20 5 * 5 = 25 # 1 * 6 = 6 2 * 6 = 12 3 * 6 = 18 4 * 6 = 24 5 * 6 = 30 6 * 6 = 36 # 1 * 7 = 7 2 * 7 = 14 3 * 7 = 21 4 * 7 = 28 5 * 7 = 35 6 * 7 = 42 7 * 7 = 49 # 1 * 8 = 8 2 * 8 = 16 3 * 8 = 24 4 * 8 = 32 5 * 8 = 40 6 * 8 = 48 7 * 8 = 56 8 * 8 = 64 # 1 * 9 = 9 2 * 9 = 18 3 * 9 = 27 4 * 9 = 36 5 * 9 = 45 6 * 9 = 54 7 * 9 = 63 8 * 9 = 72 9 * 9 = 81 print('打印出右下角的九九乘法表') for i in range(1, 10): for k in range(1, 10 -i): print(end = " ") for j in range(1, i+1): result = i * j print('%d * %d = %2d'%(j, i, result), end= " ") print() # 打印出右下角的九九乘法表 # 1 * 1 = 1 # 1 * 2 = 2 2 * 2 = 4 # 1 * 3 = 3 2 * 3 = 6 3 * 3 = 9 # 1 * 4 = 4 2 * 4 = 8 3 * 4 = 12 4 * 4 = 16 # 1 * 5 = 5 2 * 5 = 10 3 * 5 = 15 4 * 5 = 20 5 * 5 = 25 # 1 * 6 = 6 2 * 6 = 12 3 * 6 = 18 4 * 6 = 24 5 * 6 = 30 6 * 6 = 36 # 1 * 7 = 7 2 * 7 = 14 3 * 7 = 21 4 * 7 = 28 5 * 7 = 35 6 * 7 = 42 7 * 7 = 49 # 1 * 8 = 8 2 * 8 = 16 3 * 8 = 24 4 * 8 = 32 5 * 8 = 40 6 * 8 = 48 7 * 8 = 56 8 * 8 = 64 # 1 * 9 = 9 2 * 9 = 18 3 * 9 = 27 4 * 9 = 36 5 * 9 = 45 6 * 9 = 54 7 * 9 = 63 8 * 9 = 72 9 * 9 = 81

3.判断101 - 200 之间有多少个素数，并输出所有的素数


from math import sqrt 
h = 0 
leap = 1 # 1 代表是素数 for m in range(101, 201): k = int(sqrt(m)) for i in range(2, k+1): if(m % i ==0): leap = 0 break if(leap == 1): print('%2d'%m, end =" ") h += 1 if(h % 5 ==0): print() leap = 1 # 101 103 107 109 113 # 127 131 137 139 149 # 151 157 163 167 173 # 179 181 191 193 197 # 199

5.一球从100 米高度落下，每次落地后反弹的高度是原高度的一半；求它在第10次落地时，经过多少米？第10次反弹多高？

#第一次落地时，为100米；第二次落地时，弹回50米，又落下50米，共200米
Sn = 100.0
Hn = 100.0/ 2
for i in range(2, 11): Sn += 2* Hn Hn /= 2 print('总共经过%f米'%Sn) print('第10次反弹的高度为：%f'%Hn) # 总共经过299.609375米 # 第10次反弹的高度为：0.097656

6.猴子吃桃问题，每天吃所剩桃子总数的一半加1 个，直到第10天，只剩下 1个桃子

#采用逆向思维
x1 = 1
for day in range(9, 0, -1): x2 = (x1+1) * 2 x1 = x2 print(x2) # 1534

7.打印出特定图形

from sys import stdout 
for i in range(4): for j in range(3-i): stdout.write(' ') #右方的空格不考虑 for k in range(2*i + 1): stdout.write('*') print() # * # *** # ***** # ******* # 只需要考虑左方的空格和*出现的方程式规律， 如k， 第0行出现1个， 第1行出现3个，所以可以得到方程 y = 2x + 1 #打印相反方向 from sys import stdout for i in range(4): for j in range(i): stdout.write(' ') for k in range(-2 * i+7): stdout.write('*') print() # ******* # ***** # *** # *

8.有一组分数序列：2/1, 3/2, 5/3, 8/5, .....求出这个数列的前20 项的和

a, b = b, a+b 的区别可以参考;

#使用3种方法
a = 1.0
b = 2.0
sum = 0.0
# for i in range(1, 21): # sum += b/a # t = b # b = a + b # a = t # print(sum) # for i in range(1, 21): # sum += b/a # a, b = b, a+b # print(sum) from _functools import reduce list= [] for i in range(1, 21): a, b = b, a+b list.append(b/a) sum = reduce(lambda x, y: x+y, list) print(sum) # 32.278294788817234

9.求 1+2！+3！。。。+20！的累加和

#方法1 
sum = 0
t = 1
for n in range(1, 21): t *= n # 进行累乘了 sum += t print(sum) # 2561327494111820313 #方法2 def operation(x): s = 1 for i in range(1, x+1): s *= i return s list = range(1, 21) sum1 = sum(map(operation, list)) print(sum1) # 2561327494111820313

10.使用递归方法求阶乘10！

def fact(i):
    sum = 0 if (i == 0): sum = 1 else: sum = i * fact(i-1) return sum print(fact(10)) # 3628800

11.利用递归函数调用方式，将所输入的5个字符，以相反顺序打印出来。

str = input('请输入若干字符：')
 
def reverse(x): if x == -1: return '' else: return str[x] + reverse(x-1) print(reverse(len(str)-1))

12 求一个3*3 矩阵的对角线的和

str = input('输入数字')# 输入9 个以 ， 为分隔符的数字
str = str.split(sep=',')
print('输入的数字为：%s'%str)
a = []
sum = 0 for i in range(3): a.append([]) #数组 for j in range(3*i, 3*(i+1)): #间隔 a[i].append(int(str[j])) print(a) for i in range(3): sum += a[i][i] print(sum) # 输入的数字为：['1', '1', '1', '1', '6', '1', '1', '1', '10'] # [[1, 1, 1], [1, 6, 1], [1, 1, 10]] # 17

13.现有一个排好序的数组，输入一个数，要求按原来的顺序插入到数组中

#方法1
print()
a = [1,4,6,9,13,16,19,28,40,100] print(len(a)) #11 a.append(0) n = len(a) number = int(input("insert a new number:\n")) end = a[n -2] if number > end: a[n-1] = number else: for i in range(n-1): if a[i] > number: temp1 = a[i] a[i] = number for j in range(i + 1,n): #将数字后移 temp2 = a[j] a[j] = temp1 temp1 = temp2 break print(a) #insert a new number: #12 #[1, 4, 6, 9, 12, 13, 16, 19, 28, 40, 100] #方法2 list = [1,3,5,7,9,11,15,16,17,18,0] a1 = int(input('请输入一个数：')) list.append(a) list.sort() print(list) #方法3 if a > list[len(list) -1]: list.append(a) else: for i in range(len(list)): if a < list[i]: list.insert(i, a) break print(list)

剑指offer中的一些题目：

1.输入一个链表，按链表值从尾到头的顺序返回一个ArrayList。

# -*- coding:utf-8 -*-
# class ListNode:
#     def __init__(self, x):
#         self.val = x
# self.next = None class Solution: # 返回从尾部到头部的列表值序列，例如[1,2,3] def printListFromTailToHead(self, listNode): # write code here li = [] while listNode: li.append(listNode.val) listNode = listNode.next return li[::-1]

2.用两个栈来实现一个队列，完成队列的Push和Pop操作。队列中的元素为int类型。

class Solution:
    def __init__(self): self.stack1 = [] self.stack2 = [] def push(self, node): # write code here self.stack1.append(node) def pop(self): # return xx if len(self.stack2) > 0: return self.stack2.pop() while self.stack1: self.stack2.append(self.stack1.pop()) if len(self.stack2) > 0: return self.stack2.pop()

3.输入某二叉树的前序遍历和中序遍历的结果，请重建出该二叉树。假设输入的前序遍历和中序遍历的结果中都不含重复的数字。例如输入前序遍历序列{1,2,4,7,3,5,6,8}和中序遍历序列{4,7,2,1,5,3,8,6}，则重建二叉树并返回。

4.旋转数组的最小数字

把一个数组最开始的若干个元素搬到数组的末尾，我们称之为数组的旋转。输入一个非减排序的数组的一个旋转，输出旋转数组的最小元素。例如数组{3,4,5,1,2}为{1,2,3,4,5}的一个旋转，该数组的最小值为1。 NOTE：给出的所有元素都大于0，若数组大小为0，请返回0。

5.斐波那契数列

大家都知道斐波那契数列，现在要求输入一个整数n，请你输出斐波那契数列的第n项（从0开始，第0项为0）。

n<=39

# -*- coding:utf-8 -*-
class Solution:
    def Fibonacci(self, n): # write code here a, b = 0, 1 if n==0: return a else: for i in range(n): a, b = b, a+b return a

6.青蛙跳台阶

一只青蛙一次可以跳上1级台阶，也可以跳上2级。求该青蛙跳上一个n级的台阶总共有多少种跳法（先后次序不同算不同的结果）。算法思想：依然是斐波那切数列问题，当跳一节台阶时还有n-1个台阶，当第一次跳两个台阶时，还有n-2个台阶要跳。同样转化为斐波那契数列问题。

# -*- coding:utf-8 -*-
class Solution:
    def jumpFloor(self, number): # write code here if number <= 0: return 1 elif number <3: return number else: a, b = 1, 2 for i in range(3, number+1): a, b = b, a+b return b

有公式：可以分步骤进行

$1, n=1 \\ 2, n=2 \\ f(n-1)+f(n-2), n\ge3$

7.变态跳台阶

一只青蛙一次可以跳上1级台阶，也可以跳上2级……它也可以跳上n级。求该青蛙跳上一个n级的台阶总共有多少种跳法。

| 1 ,(n=0 )

f(n) = | 1 ,(n=1 )

| 2*f(n-1),(n>=2)

$f(1)=1, f(2)=2,f(3)=4\cdots\cdots f(n)=2*f(n-1)=2^{n-1}$

# -*- coding:utf-8 -

class Solution:
    def jumpFloorII(self, number): # write code here if number <=1: return 1 else: return 2*self.jumpFloorII(number-1)

8.矩形覆盖

我们可以用2*1的小矩形横着或者竖着去覆盖更大的矩形。请问用n个2*1的小矩形无重叠地覆盖一个2*n的大矩形，总共有多少种方法？

算法思想：当第一步是竖着放，右边还剩n-1个区域，当第一步横着放时，左下角应必须横着放，右边还剩n-2个区域，可以看出这仍斐波那切数列问题f(n)=f(n-1)+f(n-2)

# -*- coding:utf-8 -*-
class Solution:
    def rectCover(self, number): # write code here if number < 3: return number else: a, b = 1, 2 for i in range(2, number): a, b = b, a+b return b

9.求二进制 1 的个数

问题：输入一个整数，输出该数二进制表示中1的个数。其中负数用补码表示。

# -*- coding:utf-8 -*-
class Solution:
    def NumberOf1(self, n): # write code here count = 0 if n <0: n = n&0xffffffff while n: count += 1 n = n&(n-1) return count

print(bin(-6&0xFFFFFFFF)) #30个1  0b11111111111111111111111111111010

10.数字的整数次方

给定一个double类型的浮点数base和int类型的整数exponent。求base的exponent次方。



# -*- coding:utf-8 -*-
class Solution:
    def Power(self, base, exponent): result = 1 if base == 0: return 0 if exponent == 0: return 1 if exponent < 0: for i in range(-exponent): result = result * base return 1/result for i in range(exponent): result = result * base return result