Python课程学习记录_15周作业

  • 实现下列函数,并将其定义为模块insert, 即存储在insert.py中:
    def insert(aList, i):

    """aList[0:i] are in ascending order,

    i.e. aList[0] <= aList[1]<= ... <= aList[i-1].

    insert(aList, i) will insert aList[i] into proper postion and

    make aList[0:i+1] in asceding order and aList[i+1:] unchanged.

    """

    def insert(aList, i):
    	key=aList[i]
    	j=i-1
    	while j>=0 and aList[j] > key:
    		aList[j+1]=aList[j]
    		j-=1
    	aList[j+1]=key
    	return aList
    这里主要实现一个插入排序
  • 二. 请使用递归实现以下函数,并将其定义为模块seqrec, 即存储在seqrec.py中。

    def seq_rec(n):

    """Return the nth term of the sequence defined by a0=1, a1=2,a2 = 2, an = an-1+an-2+an-3,

    i.e., seq_rec(n) returns an.

    """

    def seq_rec(n):
    	if n == 0: 
    		return 1
    	elif n == 1 or n == 2:
    		return 2
    	else:
    		return seq_rec(n-1)+seq_rec(n-2)+seq_rec(n-3)

    三. 请使用迭代(非递归)实现以下函数,并将其定义为模块seqiter, 即存储在seqiter.py中。

    def seq_iter(n):

    """Return the nth term of the sequence defined by a0=1, a1=2,a2 = 2, an = an-1+an-2+an-3,

    扫描二维码关注公众号,回复: 2288411 查看本文章

    i.e., seq_iter(n) returns an.

    """实现

    def seq_iter(n):
    	a=[1,2,2]
    	if n == 0: 
    		return a[0]
    	elif n == 1 or n == 2:
    		return a[1]
    	else:
    		i=3
    		for i in range(3,n+1):
    			temp=a[i-1]+a[i-2]+a[i-3]
    			a.append(temp)
    		return a[n]

猜你喜欢

转载自blog.csdn.net/cyanchen666/article/details/80674971