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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，我似乎找不到破坏它的输入。也许是因为我一直盯着代码太久了，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!"
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1 回答 1

0

您的原件不太清晰，所以首先我使它面向对象： MedianHeapq支持方法rebalance(), add(), remove(), size(), median()。我们非常想minset,maxset从客户端代码中隐藏成员，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，你只需编写一个访问器。我们还添加了一种__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[i]到处索引，将s,x数组重命名为op,val，在 上添加提示，raw_input()在输入阶段拒绝无效操作。你对中位数的实际计算让我感到困惑（你什么时候想要浮点数，什么时候是整数？这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。也使它可测试。您可能在代码中出现符号错误，我不知道，我稍后再看。接下来，我们将希望通过编写一些 PyUnit 测试用例来自动化它。doctest 也是一种可能。待定。

好的，我想我看到了关于定位中位数的草率的错误。请记住，minset 和 maxset 的大小不匹配可能为 +/-1。因此，请更加注意中位数的确切位置。

#!/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__()
于 2012-08-26T05:13:06.930 回答