<p>让我们消除这种疑虑:
<strong>第二个版本</strong>稍快:<strong>列表理解更快</strong>,但两个数组循环和同样多的条件条件在一次迭代中被丢弃。在</p>
<pre><code>def kth_order_statistic1(array,k):
pivot = (array[0] + array[len(array) - 1]) // 2
l = [x for x in array if x < pivot]
m = [x for x in array if x == pivot]
r = [x for x in array if x > pivot]
if k <= len(l):
return kth_order_statistic1(l, k)
elif k > len(l) + len(m):
return kth_order_statistic1(r, k - len(l) - len(m))
else:
return m[0]
def kth_order_statistic2(array,k):
pivot = (array[0] + array[len(array) - 1]) // 2
l = []
m = []
r = []
for x in array:
if x < pivot:
l.append(x)
elif x > pivot:
r.append(x)
else:
m.append(x)
if k <= len(l):
return kth_order_statistic2(l, k)
elif k > len(l) + len(m):
return kth_order_statistic2(r, k - len(l) - len(m))
else:
return m[0]
def kth_order_statistic3(array,k):
pivot = (array[0] + array[len(array) - 1]) // 2
l = []
m = []
r = []
for x in array:
if x < pivot: l.append(x)
for x in array:
if x== pivot: m.append(x)
for x in array:
if x > pivot: r.append(x)
if k <= len(l):
return kth_order_statistic3(l, k)
elif k > len(l) + len(m):
return kth_order_statistic3(r, k - len(l) - len(m))
else:
return m[0]
import time
import random
if __name__ == '__main__':
A = range(100000)
random.shuffle(A)
k = random.randint(1, len(A)-1)
start_time = time.time()
for x in range(1000) :
kth_order_statistic1(A,k)
print("--- %s seconds ---" % (time.time() - start_time))
start_time = time.time()
for x in range(1000) :
kth_order_statistic2(A,k)
print("--- %s seconds ---" % (time.time() - start_time))
start_time = time.time()
for x in range(1000) :
kth_order_statistic3(A,k)
print("--- %s seconds ---" % (time.time() - start_time))
</code></pre>
<p><br/></p>
^{pr2}$
<p>时间可能因随机抽签而不同,但三者之间的差异几乎相同。在</p>