python - Python 中的 Inteviewstreet 中位数。除了第一个测试用例之外的所有测试都失败

Question

所以我写了这段代码，它通过了第一个测试用例，其余的都失败了。但是，我似乎找不到打破它的输入。也许是因为我盯着代码看的时间太长了，但我会很感激任何帮助。 该算法对当前列表的最小和最大一半使用两个优先级队列。这是代码:

#!/bin/python
import heapq

def fix(minset, maxset):
    if len(maxset) > len(minset):
        item = heapq.heappop(maxset)
        heapq.heappush(minset, -item)
    elif len(minset) > (len(maxset) + 1):
        item = heapq.heappop(minset)
        heapq.heappush(maxset, -item)

N = int(raw_input())

s = []
x = []

for i in range(0, N):

        tmp = raw_input()
        a, b = [xx for xx in tmp.split(' ')]
        s.append(a)
        x.append(int(b))

minset = []
maxset = []

for i in range(0, N):
    wrong = False
    if s[i] == "a":
        if len(minset) == 0:
            heapq.heappush(minset,-x[i])
        else:
            if x[i] > minset[0]:
                heapq.heappush(maxset, x[i])
            else:
                heapq.heappush(minset, -x[i])
        fix(minset, maxset)
    elif s[i] == "r":
        if -x[i] in minset:
            minset.remove(-x[i])
            heapq.heapify(minset)
        elif x[i] in maxset:
            maxset.remove(x[i])
            heapq.heapify(maxset)
        else:
            wrong = True
        fix(minset, maxset)
    if len(minset) == 0 and len(maxset) == 0:
        wrong = True

    if wrong == False:
        #Calculate median
        if len(minset) > len(maxset):
            item = - minset[0]
            print int(item)
        else:
            item = ((-float(minset[0])) + float(maxset[0])) / 2
            if item.is_integer():
                print int(item)
                continue
            out =  str(item)
            out.rstrip('0')
            print out
    else:
        print "Wrong!"

Answer 1

你的原文不是那么清晰，所以首先我让它面向对象: MedianHeapq 支持方法 rebalance()、add()、remove()、size()、median()。我们非常想从客户端代码中隐藏成员 minset,maxset，出于各种合理的原因:防止客户端交换它们、修改它们等。如果客户端需要看到它们，您只需编写一个访问器. 我们还添加了一个 __str__() 方法，我们将使用该方法进行可视化调试，让您的生活更轻松。

还添加了易读性更改以避免到处使用 [i] 进行索引，将 s,x 数组重命名为 op,val，添加提示在 raw_input() 上，拒绝输入阶段的无效操作。 你对中位数的实际计算让我很困惑(你什么时候需要 float ，什么时候需要整数？rstrip('0') 有点古怪)，所以我重写了它，如果你想要别的东西，就改变它. 该算法的讨论是 here .

现在它清晰易读且独立。也使其可测试。 您可能在代码中犯了符号错误，我不知道，我稍后再看。 接下来我们要通过编写一些 PyUnit 测试用例来自动化它。 doctest 也是一种可能性。待定。

好吧，我想我看到了关于定位中位数的草率错误。请记住，minset 和 maxset 的大小可能不匹配 +/-1。因此，请更加注意中位数的准确位置。

#!/bin/python
import heapq    

class MedianHeapq(object):
    def __init__(self):
        self.minset = []
        self.maxset = []        
    def rebalance(self):
        size_imbalance = len(self.maxset) - len(self.minset)
        if len(self.maxset) > len(self.minset):
        #if size_imbalance > 0:
            item = heapq.heappop(self.maxset)
            heapq.heappush(self.minset, -item)
        #elif size_imbalance < -1:
        elif len(self.minset) > (len(self.maxset) + 1):
            item = heapq.heappop(self.minset)
            heapq.heappush(self.maxset, -item)
    def add(self, value, verbose=False):
        if len(self.minset) == 0:
            heapq.heappush(self.minset,-value)
        else:
            if value > self.minset[0]:
                heapq.heappush(self.maxset, value)
            else:
                heapq.heappush(self.minset, -value)
        self.rebalance()
        if verbose: print self.__str__()
        return False
    def remove(self,value,verbose=False):
        wrong = False
        if -value in self.minset:
            minset.remove(-value)
            heapq.heapify(self.minset)
        elif value in maxset:
            maxset.remove(value)
            heapq.heapify(self.maxset)
        else:
            wrong = True
        self.rebalance()
        if verbose: print self.__str__()
        return wrong
    def size(self):
        return len(self.minset)+len(self.maxset)
    def median(self):
        if len(self.minset) > len(self.maxset):
            item = - self.minset[0]
            return int(item)
        else:
            item = (-self.minset[0] + self.maxset[0]) / 2.0
            # Can't understand the intent of your code here: int, string or float?
            if item.is_integer():
                return int(item)
            #    continue # intent???
            else:
               return item
            # The intent of this vv seems to be round floats and return '%.1f' % item ?? 
            #out =  str(item)
            #out.rstrip('0') # why can't you just int()? or // operator?
            #return out


    def __str__(self):
        return 'Median: %s Minset:%s Maxset:%s' % (self.median(), self.minset,self.maxset)

# Read size and elements from stdin
N = int(raw_input('Size of heap? '))
op = []
val = []
while(len(val)<N):
        tmp = raw_input('a/r value : ')
        op_, val_ = tmp.split(' ')
        if op_ not in ['a','r']: # reject invalid ops
            print 'First argument (operation) must be a:Add or r:Remove! '
            continue
        op.append(op_)
        val.append(int(val_))

mhq = MedianHeapq()
for op_,val_ in zip(op,val): # use zip to avoid indexing with [i] everywhere
    wrong = False
    if op_ == 'a':
        wrong = mhq.add(val_)
    elif op_ == 'r':
        wrong = mhq.remove(val_)

    assert (mhq.size()>0), 'Heap has zero size!'

    assert (not wrong), 'Heap structure is wrong!'

    if not wrong:
        print mhq.__str__()