In terms of time complexity, both methods have the same linear time complexity O(n). However, the first method (max(arr))might have a slight advantage in terms of performance, as it uses a single function optimized for this task, whereas the second method involves a loop with multiple comparisons in each iteration. The actual performance difference might be negligible for small lists, but for larger lists, the first method might be slightly faster due to its optimized nature.

因此,你更喜欢使用“<代码>max(r)功能”,因为功能是为这一特定目的设计的,而且很可能是完全优化的。

Times for 100000 elements (random, increasing or decreasing) with your two ways and one without using max at all:

random
  1.79 ± 0.02 ms  selfmade
  2.04 ± 0.02 ms  just_max
 17.62 ± 0.50 ms  repeated_max

increasing
  1.89 ± 0.01 ms  selfmade
  2.09 ± 0.04 ms  just_max
 17.23 ± 0.09 ms  repeated_max

decreasing
  1.76 ± 0.04 ms  selfmade
  2.06 ± 0.03 ms  just_max
 18.11 ± 0.46 ms  repeated_max

Python: 3.11.4 (main, Jun 24 2023, 10:18:04) [GCC 13.1.1 20230429]

图瓦卢

def just_max(arr):
    x = max(arr)
    return x

def repeated_max(arr):
    x = 0
    for ele in arr:
        x = max(x,ele)
    return x

def selfmade(arr):
    x = 0
    for ele in arr:
        if ele > x:
            x = ele
    return x


funcs = just_max, repeated_max, selfmade

import random
from timeit import timeit
from statistics import mean, stdev
import sys

def test(label, arr):
    print(label)
    times = {f: [] for f in funcs}
    def stats(f):
        ts = [t * 1e3 for t in sorted(times[f])[:5]]
        return f {mean(ts):6.2f} ± {stdev(ts):4.2f} ms  

    for _ in range(25):
        for f in funcs:
            t = timeit(lambda: f(arr), number=1)
            times[f].append(t)

    for f in sorted(funcs, key=stats):
        print(stats(f), f.__name__)

    print()

n = 10**5
test( random , random.choices(range(10**9), k=n))
test( increasing , list(range(n)))
test( decreasing , list(range(n)[::-1]))

print( Python: , sys.version)

在业绩和复杂性方面,要打破两种方法:

<<>Method 1: Usingmax(arr)

这种方法使用<条形码>最大()的沙捞越功能,在内部通过该清单进行调节,并跟踪所达到的最高值。 它为找到最高价值提供了简明和可读的方法。

• Time Complexity: O(n), where n is the number of elements in the list. The max() function needs to go through each element of the list once to find the maximum.
• Space Complexity: O(1), because only a constant amount of additional memory is required to keep track of the maximum value.

<>Method2:

这种方法通过清单进行人工调节,并将每个要素与现有最高值进行比较。 如果现有要素更大,则更新上限。

• Time Complexity: O(n), same as the previous method. It iterates through each element once and performs a constant-time comparison.
• Space Complexity: O(1), similar to the previous method. Only a few extra variables are used.

这两种方法都具有同样的复杂性,但是在“顶端”(max(>)功能中,可能有一些最佳业绩,由于在C实施,实际上可能使其稍快。 然而,如你提供的小型清单,差异可能并不明显。

In terms of which one is preferred, using max(arr) is generally recommended due to its simplicity, readability, and potential performance optimizations. It s a clear and concise way to express the intention of finding the maximum value in the list. The manual comparison method might be useful in specific cases where you need more control over the comparison process or if you want to perform additional operations during the iteration.

总之,这两种方法在时间方面和空间的复杂性上都相当可观,以便在清单中找到最高价值。 <代码>max()功能通常是首选功能,除非有具体理由人工操作和比较。 手册方法的比较数量与内在功能相比,没有显著影响时间的复杂性。