数据结构列表练习

Posted 2020-10-29

数据结构列表练习

1、求10万内的所有素数
    import datetime
    start=datetime.datetime.now()
    import math
    primenumber = [2]
    flag = False
    for i in range(3,100000,2):
        if i > 10 and i % 10 == 5:
            continue
        edge = math.ceil(math.sqrt(i))
        for j in primenumber:
            if i % j == 0 :
                flag = True
                break
            if j >= edge:
                flag = False
                break
        if not flag:
            print(i)
            primenumber.append(i)
    delta = (datetime.datetime.now() - start).total_seconds() #时间计算
    print(delta)

2、计算杨辉三角前6行
（1）方法1：补头和尾，中间循环
    triangle = [[1],[1,1]]
    n = 6
    for i in range(2,n):
        newline = [1]
        pre = trangle[i-1]
        for j in range(i-1):
            val = pre[j] + pre[j+1]
            newline.append(val)
        newline.append(1)
        trangle.append(newline)
    print(trangle)

（2）方法1变体
    n = 6
    for i in range(n): #如果不想每次判断，则把[1]放在循环外边
        newline = [1]
        if i == 0:
            print(newline)
            continue
        for j in range(i-1):
            val = pre[j] + pre[j+1]
            newline.append(val)
        newline.append(1)
        print(newline)
        pre = newline

（3）方法2：尾部补0法
    n = 6
    for i in range(1,n):
        if i == 0:
            newline = [1]
            print(newline)
            newline.append(0)
            oldline = newline
            continue            
        newline = []
        for j in range(i+1):
            newline.append(oldline[j-1] + oldline[j])
        print(newline)
        newline.append(0)
        oldline = newline

（4）方法2变体：尾部补0法
    n = 6
    newline = [1]
    print(newline)
    newline.append(0)
    for i in range(1,n):
        newline = []
        for j in range(i+1):
            newline.append(oldline[j-1] + oldline[j])
        print(newline)
        newline.append(0)
        oldline = newline

（5）方法3：一次性开辟空间，再对称。
    n = 6
    triangle = []
    for i in range(n):
        row = [1] * (i+1) #一次性开辟
        triangle.append(row)
        for j in range(1,i//2+1): # i=2,第三行才能进来
            val = triangle[i-1][j-1] + triangle[i-1][j]
            row[j] = val
            if i != 2*j: #奇数个数的中点跳过
                row[-j-1] = val
    print(triangle)

（6）方法4：单行覆盖
    n = 6
    row = [1] * n #一次性开辟足够的空间
    for i in range(n):
        offset = n - i
        z = 1 #因为会有覆盖影响计算，所以引入一个临时变量
        for j in range(1,i//2+1):
            val = z + row[j]
            row[j],z = val,row[j]
            if i != 2*j:
                row[-j-offset] = val
        print(row[:i+1])

3、依次接收用户输入的3个数，排序后打印
（1）分支结构
    nums = []
    out = None
    for i in range(3):
        nums.append(int(input('{}: '.format(i))))
    if nums[0]>nums[1]:
        if nums[1]>nums[2]:
            out = [0,1,2]
        elif nums[0]>nums[2]:
            out = [0,2,1]
        else:
            out = [2,0,1]
    elif nums[0]>nums[2]:
        out = [1,0,2]
    elif nums[1]>nums[2]:
        out = [1,2,0]
    else:
        out = [2,1,0]
    out.reverse()
    for i in out:
        print(nums[i],end=',')

（2）max函数
    nums = []
    out =  None
    for i in range(3):
        nums.append(int(input('{}: '.format(i))))
    #此处不能使用for循环，不能一边迭代该列表，同时删除或者增加该列表
    while True:
        cur = max(nums)
        print(cur)
        nums.remove(cur)
        if len(nums) == 1:
            print(nums[0])
            break

（3）列表sort实现   
    nums = []
    for i in range(3):
        nums.append(int(input('{}: '.format(i))))
    nums.sort()
    print(nums)

4、经典冒泡法。
（1）初始实现
    lst = [9,5,1,2,6]
    length = len(lst)
    for i in range(length):
        for j in range(length-1-i):
            if lst[j] > lst[j+1]:
                lst[j], lst[j+1] = lst[j+1], lst[j]
    print(lst)

（2）优化实现
    lst = [9,5,1,2,6]
    length = len(lst)
    for i in range(length):
        flag = False
        for j in range(length-1-i):
            if lst[j] > lst[j+1]:
                lst[j], lst[j+1] = lst[j+1], lst[j]
                flag = True
        if not flag:
            break
    print(lst)