问题
I was wondering if there is a way to find min & max of a list without using min/max functions in Python. So i wrote a small code for the same using recursion. My logic is very naive: I make two stacks (min_stack and max_stack) which keep track of minimum and maximum during each recursive call. I have two questions:
- Could somebody help me estimate the complexity of my code?
- Is there a better way to do this? Will sorting the list using mergesort/quicksort and picking up first and last element give a better performance?
Thank you
Here is my attempt in Python:
minimum = []
maximum = []
# Defining Stack Class
class Stack:
def __init__(self) :
self.items = []
def push(self, item) :
self.items.append(item)
def pop(self) :
return self.items.pop()
def access(self, index):
return self.items[index]
def isEmpty(self) :
return (self.items == [])
def length(self):
return len(self.items)
def minmax(input_list):
# make two stacks, one for min and one for max
min_stack = Stack()
max_stack = Stack()
# comparing the first two elements of the list and putting them in appropriate stack
if input_list[0]<input_list[1]:
min_stack.push(input_list[0])
max_stack.push(input_list[1])
else:
max_stack.push(input_list[0])
min_stack.push(input_list[1])
# Pushing remaining elements of the list into appropriate stacks.
for i in range(2, len(input_list)):
if input_list[i] < min_stack.access(-1):
min_stack.push(input_list[i])
else:
max_stack.push(input_list[i])
# to find minimum
minlist = []
while min_stack.length() > 0:
minlist.append(min_stack.pop())
# to find maximum
maxlist = []
while max_stack.length() > 0:
maxlist.append(max_stack.pop())
if len(minlist) > 1:
minmax(minlist)
else:
minimum.append(minlist)
if len(maxlist) > 1:
minmax(maxlist)
else:
maximum.append(maxlist)
def main():
input_list = [2, 0, 2, 7, 5, -1, -2]
print 'Input List is: ', input_list
minmax(input_list)
print 'Global Minimum is: ', minimum[0]
print 'Global Maximum is: ', maximum[len(maximum)-1]
if __name__ == "__main__":
main()
回答1:
Using sorted() would, of course, be reliable, quick to write, and high performance for moderate-sized lists because it is built-in. For large lists, an O(n) algorithm would be faster e.g.:
def minmax1 (x):
# this function fails if the list length is 0
minimum = maximum = x[0]
for i in x[1:]:
if i < minimum:
minimum = i
else:
if i > maximum: maximum = i
return (minimum,maximum)
print(minmax1([9,8,7,6,5,4,3,2,1,11,12,13,14,15,16,17,18,19]))
print(minmax1([1]))
print(minmax1([2, 0, 2, 7, 5, -1, -2]))
... for which the output is:
(1, 19)
(1, 1)
(-2, 7)
I was interested to check the performance of the two alternatives. On my PC running Windows XP and Python 3.2.3, I found that the sorting approach is faster than the minmax1() function defined above for lists of fewer than 500 elements but, for longer lists, the O(n) minmax1() is faster. My timing test code was as follows:
def minmax_sort(x):
x = sorted(x)
return (x[0],x[-1])
import timeit
aa = list(range(0,100))
a = aa
while (1):
stime = min(timeit.repeat('minmax_sort(a)', "from __main__ import minmax_sort,a",number=1000))
mtime = min(timeit.repeat('minmax1(a)', "from __main__ import minmax,a",number=1000))
if (stime > mtime):
break
else:
a = a + aa
print(len(a))
回答2:
Finding Min and Max of a list using only recursion.
I had this similar assignment last week and I divided the code into three parts.
Step 1: Finding the minimum value in list
def RecursiveMin(L):
if len(L)==2:
if L[0]<L[1]:
return L[0]
else:
return L[1]
else:
X= RecursiveMin(L[1:])
if L[0]<X:
return L[0]
else:
return X
Step 2: Sorting the list using into ascending order (smallest to largest)
def Sort(x):
L=sorted(x)
if x==L:
return x
else:
i=0
for i in range (len(x)):
if x[i] > x[i+1] :
break
unsortedPart = x[i:]
R = RecursiveMin(unsortedPart)
I = unsortedPart.index(R)
for j in range (len(x)):
if x[j] > R :
del x[(I+i)]
x.insert(j,R)
break
return Sort(x)
(I have previously answered a sorting a list question and provided the same code. So please don't flag me for plagiarism since it's my own code Likn: Finding minimum element in a list (recursively) - Python ).
**Step 3: Make a new function with an argument whether the user wants Min value or Max*
def minMax(lst,user):
if user == min:
return Sort(lst)
elif user == max:
s = Sort(lst)
return s[::-1]
The last step is not recursive but if you compile all the three steps into one function it will be "1 recursive function". P.S if your question was only about finding the Min and Max in a list you can skip Step 2 and make a few changes to to Step 1 and Step 3
回答3:
If you use the sorted() function, just call the first index for minimum and the last index for maximum. No need for a for loop.
def minimum(x):
x = sorted(x)
return x[0]
def maximum(x):
x = sorted(x)
return x[-1]
print(minimum([2, -5, 79, 20, -67])
print(maximum([45, -78, 950, 39, -567])
The output is:
-67
950
回答4:
This would be very simple and easy to understand. Hope this will help you.
arr = []
num = int(input("Enter number of elements in list: "))
for i in range(0, num):
ele = int(input("Enter elements: "))
arr.append(ele)
min = arr[ 0 ]
for a in arr:
if a < min:
min = a
print ("The minimum number in the list is: ", min)
max = arr[0]
for a in arr:
if a > max:
max = a
print("The maximum number in the lit is: ", max)
回答5:
I was asked to implement it using Stacks as a way of exercise.
I'm surprised several of the solutions require multiple passes through the list to determine the minimum and maximum. Here's a simple Python 3 recursive solution, using stacks per the OP, that only makes one pass through the list:
def minmax(array, minimum=None, maximum=None):
head, *tail = array
if minimum is None:
minimum = [head]
elif head < minimum[-1]:
minimum.append(head)
if maximum is None:
maximum = [head]
elif head > maximum[-1]:
maximum.append(head)
if tail:
return minmax(tail, minimum, maximum)
return minimum.pop(), maximum.pop()
if __name__ == "__main__":
array = [2, 0, 2, 7, 5, -1, -2]
minimum, maximum = minmax(array)
print(array, minimum, maximum)
array = [13]
minimum, maximum = minmax(array)
print(array, minimum, maximum)
array = [9, 8, 7, 6, 5, 4, 3, 2, 1, 11, 12, 13, 14, 15, 16, 17, 18, 19]
minimum, maximum = minmax(array)
print(array, minimum, maximum)
Though the stacks aren't strickly necessary for this code to work.
来源：https://stackoverflow.com/questions/15148491/min-and-max-of-a-list-in-python-without-using-min-max-function