CS61A_Lab_05

Q1

题目描述：

代码实现：

def coords(fn, seq, lower, upper):
    """
    >>> seq = [-4, -2, 0, 1, 3]
    >>> fn = lambda x: x**2
    >>> coords(fn, seq, 1, 9)
    [[-2, 4], [1, 1], [3, 9]]
    """
    "*** YOUR CODE HERE ***"
    #seq序列中的数字经过函数变化之后要在lower和upper的区间之中，否则就会被舍弃
    #输出是以list的形式存在的
    #所以我可以使用recursion
    if not seq:
      return []
    if fn(seq[0])>=lower and fn(seq[0])<=upper:
      return [[seq[0],fn(seq[0])]]+coords(fn,seq[1:],lower,upper)
    else:
      return coords(fn,seq[1:],lower,upper) 
    #return [[x,f(x)] for x in seq if fn(x)>=lower and fn(x)<=upper]

分析：

别人的实现非常的简洁，以后我也可以这么做。

我的实现好像有漏洞，当list只有一个了的时候，seq[1:]会超出list范围，可能会引起报错。

 

Q2：

题目描述：

代码实现：

def riffle(deck):
    """Produces a single, perfect riffle shuffle of DECK, consisting of
    DECK[0], DECK[M], DECK[1], DECK[M+1], ... where M is position of the
    second half of the deck.  Assume that len(DECK) is even.
    >>> riffle([3, 4, 5, 6])
    [3, 5, 4, 6]
    >>> riffle(range(20))
    [0, 10, 1, 11, 2, 12, 3, 13, 4, 14, 5, 15, 6, 16, 7, 17, 8, 18, 9, 19]
    """
    "*** YOUR CODE HERE ***"
    M=len(deck)//2
    return [deck[k//2 + M*(k%2)] for k in range(len(deck))] 

分析：

这个代码实现很巧妙。

第一个亮点就是 for k in range(len(deck)) 来进行记数，为后面判断奇偶做好准备。

第二个亮点就是注意到 10 11 12 13…… 其实就是0 1 2 3 加上中间值。用k//2来重复两次，太妙了。

Q3：

题目描述：

 

 代码：

def sprout_leaves(t,leaves):
    """Sprout new leaves containing the data in leaves at each leaf in
    the original tree t and return the resulting tree.

    >>> t1 = tree(1, [tree(2), tree(3)])
    >>> print_tree(t1)
    1
      2
      3
    >>> new1 = sprout_leaves(t1, [4, 5])
    >>> print_tree(new1)
    1
      2
        4
        5
      3
        4
        5

    >>> t2 = tree(1, [tree(2, [tree(3)])])
    >>> print_tree(t2)
    1
      2
        3
    >>> new2 = sprout_leaves(t2, [6, 1, 2])
    >>> print_tree(new2)
    1
      2
        3
          6
          1
          2
    """
    "*** YOUR CODE HERE ***"
    if is_leaf(t):
        return tree(label(t), [tree(x) for x in leaves])
        #增加枝叶
    else:
        return tree(label(t), [sprout_leaves(branch, leaves) for branch in branches(t)])

 Q6 add tree 树的合并

代码：

def add_trees(t1, t2):
    """
    >>> numbers = tree(1,
    ...                [tree(2,
    ...                      [tree(3),
    ...                       tree(4)]),
    ...                 tree(5,
    ...                      [tree(6,
    ...                            [tree(7)]),
    ...                       tree(8)])])
    >>> print_tree(add_trees(numbers, numbers))
    2
      4
        6
        8
      10
        12
          14
        16
    >>> print_tree(add_trees(tree(2), tree(3, [tree(4), tree(5)])))
    5
      4
      5
    >>> print_tree(add_trees(tree(2, [tree(3)]), tree(2, [tree(3), tree(4)])))
    4
      6
      4
    >>> print_tree(add_trees(tree(2, [tree(3, [tree(4), tree(5)])]), \
    tree(2, [tree(3, [tree(4)]), tree(5)])))
    4
      6
        8
        5
      5
    """
    "*** YOUR CODE HERE ***"
    if is_leaf(t1):
      return tree(label(t1)+label(t2),branches(t2))
    elif is_leaf(t2):
      return tree(label(t2)+label(t1),branches(t1))
    else:
      fewer_branch_t,more_branch_t=sorted([branches(t1),branches(t2)],key=len)
      #sorted()函数将该列表中的两个子节点列表按照长度进行排序，并返回一个新的列表
      #新的列表中第一个元素是子节点较少的树的子节点列表
      # 第二个元素是子节点较多的树的子节点列表。每个branches只会有两个元素，len<=2


      pad_t1=fewer_branch_t+[tree(0) for i in range(len(more_branch_t)-len(fewer_branch_t))]
      pad_t2=more_branch_t
      #是为了让两个list 元素对应，方便后面使用zip连接，再进行递归
      return tree(label(t1)+label(t2),\
                  [add_trees(b1,b2) for b1, b2 in zip(pad_t1,pad_t2)])