我能想到的最快的方法是对原始列表进行排序，然后选择第一个和最后一个元素。这样可以避免多次循环，但是会破坏列表的原始结构。这可以通过简单地复制列表并仅对复制的列表进行排序来解决。我很好奇这是否比使用此快速示例脚本仅使用max（）和min（）更慢：

import time

l = [1,2,4,5,3]

print "Run 1"
t1 = time.time()
print "Min =", min(l)
print "Max =", max(l)
print "time =", time.time() - t1
print ""
print "l =", l
print ""


l = [1,2,4,5,3]
l1 = list(l)

print "Run 2"
t1 = time.time()
l1.sort()
print "Min =", l1[0]
print "Max =", l1[-1]
print "time =", time.time() - t1
print ""
print "l =", l
print "l1 =", l1
print ""


l = [1,2,4,5,3]

print "Run 3"
minimum = float('inf')
maximum = float('-inf')
for item in l:
    if item < minimum:
        minimum = item
    if item > maximum:
        maximum = item
print "Min =", minimum
print "Max =", maximum
print "time =", time.time() - t1
print ""
print "l =", l

出乎意料的是，第二种方法在我的计算机上速度提高了约10ms。不确定使用非常大的列表会产生多大的效果，但是这种方法至少对于您提供的示例列表而言更快。

我在计时脚本中添加了@Martijn Pieters的简单循环算法。（因为计时是值得在此问题中探讨的唯一重要参数。）我的结果是：

Run 1: 0.0199999809265s
Run 2: 0.00999999046326s
Run 3: 0.0299999713898s

编辑：包括timeit模块进行计时。

import timeit
from random import shuffle

l = range(10000)
shuffle(l)

def Run_1():
    #print "Min =", min(l)
    #print "Max =", max(l)
    return min(l), max(l)

def Run_2():
    l1 = list(l)
    l1.sort()
    #print "Min =", l1[0]
    #print "Max =", l1[-1]
    return l1[0], l1[-1]


def Run_3():
    minimum = float('inf')
    maximum = float('-inf')
    for item in l:
        if item < minimum:
            minimum = item
        if item > maximum:
            maximum = item
    #print "Min =", minimum
    #print "Max =", maximum
    return minimum, maximum


if __name__ == '__main__':
    num_runs = 10000
    print "Run 1"
    run1 = timeit.Timer(Run_1)
    time_run1 = run1.repeat(3, num_runs)
    print ""
    print "Run 2"
    run2 = timeit.Timer(Run_2)
    time_run2 = run2.repeat(3,num_runs)
    print ""
    print "Run 3"
    run3 = timeit.Timer(Run_3)
    time_run3 = run3.repeat(3,num_runs)
    print ""

    print "Run 1"
    for each_time in time_run1:
        print "time =", each_time
    print ""
    print "Run 2"
    for each_time in time_run2:
        print "time =", each_time
    print ""
    print "Run 3"
    for each_time in time_run3:
        print "time =", each_time
    print ""

我的结果是：

Run 1
time = 3.42100585452
time = 3.39309908229
time = 3.47903182233

Run 2
time = 26.5261287922
time = 26.2023346397
time = 26.7324208568

Run 3
time = 3.29800945144
time = 3.25067545773
time = 3.29783778232

对于大型数组，sort算法非常慢。