标签：return Tree rest cs61a dis08 Link 2021 lnk first

网址 https://inst.eecs.berkeley.edu/~cs61a/fa21/disc/disc08/
problem1：

problem2：

problem3:
递归

    def sum_nums(s):
        """
        >>> a = Link(1, Link(6, Link(7)))
        >>> sum_nums(a)
        14
        """
        "*** YOUR CODE HERE ***"
        if s is Link.empty:
            return 0
        return s.first + sum_nums(s.rest)

problem4:
这个也是用递归，因为不好判断哪一个链表最短，所以直接迭代比较麻烦

    def multiply_lnks(lst_of_lnks):
        """
        >>> a = Link(2, Link(3, Link(5)))
        >>> b = Link(6, Link(4, Link(2)))
        >>> c = Link(4, Link(1, Link(0, Link(2))))
        >>> p = multiply_lnks([a, b, c])
        >>> p.first
        48
        >>> p.rest.first
        12
        >>> p.rest.rest.rest is Link.empty
        True
        """
        # Implementation Note: you might not need all lines in this skeleton code
        product = 1
        for lnk in lst_of_lnks:
            if lnk is Link.empty:
                return Link.empty
            product *= lnk.first
    	lst_of_lnks_rset = [lnk.rest for lnk in lst_of_lnks]
        return Link(product, multiply_lnks(lst_of_lnks_rset))

problem5:
递归和迭代

    def flip_two(s):
        """
        >>> one_lnk = Link(1)
        >>> flip_two(one_lnk)
        >>> one_lnk
        Link(1)
        >>> lnk = Link(1, Link(2, Link(3, Link(4, Link(5)))))
        >>> flip_two(lnk)
        >>> lnk
        Link(2, Link(1, Link(4, Link(3, Link(5)))))
        """
        "*** YOUR CODE HERE ***"
        #if s is Link.empty or s.rest is Link.empty:
        #    return
        #tmp = s.rest.first
        #s.rest.first = s.first
        #s.first = tmp
        #flip_two(s.rest.rest)
        # For an extra challenge, try writing out an iterative approach as well below!
        "*** YOUR CODE HERE ***"
        while s is not Link.empty and s.rest is not Link.empty:
            tmp = s.rest.first
            s.rest.first = s.first
            s.first = tmp
            s = s.rest.rest

problem6:

    def make_even(t):
        """
        >>> t = Tree(1, [Tree(2, [Tree(3)]), Tree(4), Tree(5)])
        >>> make_even(t)
        >>> t.label
        2
        >>> t.branches[0].branches[0].label
        4
        """
        "*** YOUR CODE HERE ***"
        if t.label % 2 == 1:
            t.label += 1
        for i in t.branches:
            make_even(i)

problem7:

    def leaves(t):
        """Returns a list of all the labels of the leaf nodes of the Tree t.

        >>> leaves(Tree(1))
        [1]
        >>> leaves(Tree(1, [Tree(2, [Tree(3)]), Tree(4)]))
        [3, 4]
        """
        "*** YOUR CODE HERE ***"
        ans = []
        if t.is_leaf():
            return [t.label]
        for i in t.branches:
            ans += leaves(i)
        return ans

problem8:

    def find_paths(t, entry):
        """
        >>> tree_ex = Tree(2, [Tree(7, [Tree(3), Tree(6, [Tree(5), Tree(11)])]), Tree(1, [Tree(5)])])
        >>> find_paths(tree_ex, 5)
        [[2, 7, 6, 5], [2, 1, 5]]
        >>> find_paths(tree_ex, 12)
        []
        """

        paths = []
        if t.label == entry:
            [[t.label]]
        for i in t.branches:
            path = find_paths(i,entry)
            for j in path:
                j = [t.label] + j
                paths.append(j)
        return paths

标签：return,Tree,rest,cs61a,dis08,Link,2021,lnk,first
来源： https://www.cnblogs.com/echoT/p/16127141.html

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