打印三角型的练习

编写函数，接受一个n，n为正整数，左右两种打印方式。要求数字必须对齐

12 11 10 9 8 7 6 5 4 3 2 1
   11 10 9 8 7 6 5 4 3 2 1
      10 9 8 7 6 5 4 3 2 1
         9 8 7 6 5 4 3 2 1
           8 7 6 5 4 3 2 1
             7 6 5 4 3 2 1
               6 5 4 3 2 1
                 5 4 3 2 1
                   4 3 2 1
                     3 2 1
                       2 1
                         1
                         1
                       2 1
                     3 2 1
                   4 3 2 1
                 5 4 3 2 1
               6 5 4 3 2 1
             7 6 5 4 3 2 1
           8 7 6 5 4 3 2 1
         9 8 7 6 5 4 3 2 1
      10 9 8 7 6 5 4 3 2 1
   11 10 9 8 7 6 5 4 3 2 1
12 11 10 9 8 7 6 5 4 3 2 1


补空格法

def trangle2(n):
    nums = [i for i in range(n,0,-1)]
    length = len(nums)
    
    for i in range(length):
        for j in range(length):
            if i<=j:
                print('{:>{}}'.format(nums[j],len(str(nums[j]))+1),end='')
            else:
                print('{}'.format(' '*(len(str(nums[j]))+1)),end='')
        print()
    for i in range(length):
        for j in range(length,0,-1):
            if i+1>=j:
                print('{:>{}}'.format(nums[-j],len(str(nums[-j]))+1),end='')
            else:
                print('{}'.format(' '*(len(str(nums[-j]))+1)),end='')
        print()

靠右对齐法

def trangle_print(n):
    nums = ' '.join(str(i) for i in range(n,0,-1))
    width = len(nums)
    start = -1
    step = 2
    points = {10**i for i in range(1,3)}
    for i in range(1,n+1):
        print('{:>{}}'.format(nums[start:],width))
        if i+1 in points:
            step += 1
        start = start -step

字符串切割法

def trangle1(n):
    nums = ' '.join([str(i) for i in range(n,0,-1)])
    length = len(nums)
    print(nums)
    for i in range(len(nums)):
        if nums[i] ==' ':
            print('{}{}'.format(' '*(i+1),nums[(i+1):]))
            
    for j in range(len(nums)-1,0,-1):
        if nums[j] == ' ':
            print('{}{}'.format(' '*(j),nums[j:]))
    print(nums)