Python：排序列表时如何停止while循环？

Question

我正在编写代码来对列表中的元素进行排序，我实现了它但无法停止 while 循环。

当列表中的所有元素都排序后，我想停止 while 循环。

代码：

a = [27,21,22,1,11,23,0]
n=len(a)
while True:
    for i in range(n-1):
        if a[i]>a[i+1]:
            temp = a[i]
            a[i] = a[i+1]
            a[i+1] = temp
    print(a)

输出：

[21, 22, 1, 11, 23, 0, 27] [21, 1, 11, 22, 0, 23, 27] [1, 11, 21, 0, 22, 23, 27] [1, 11, 0, 21, 22, 23, 27] [1, 0, 11, 21, 22, 23, 27] [0, 1, 11, 21, 22, 23, 27] [0, 1, 11, 21, 22, 23, 27] [0, 1, 11, 21, 22, 23, 27] [0, 1, 11, 21, 22, 23, 27]........

我知道排序列表上有 10 个问答，但我只想以这种方式实现（如果可能的话）。

谁能帮我停止这个while循环？

Answer 1

也许比 chepner 的效率高一点，每次交换都没有额外的分配，而是使用temp 来检测是否有变化：

a = [27,21,22,1,11,23,0]
n = len(a)
dummy = object()
while True:
    temp = dummy
    for i in range(n-1):
        if a[i]>a[i+1]:
            temp = a[i]
            a[i] = a[i+1]
            a[i+1] = temp
    if temp is dummy:
        break
    print(a)

Answer 2

为什么不只是：

a = sorted(a)

好吧，如果你想这样做：

a = [27,21,22,1,11,23,0]; a_bis = []
n=len(a)
while a_bis != a:
    a_bis = a.copy()
    for i in range(n-1):
        if a[i]>a[i+1]:
            temp = a[i]
            a[i] = a[i+1]
            a[i+1] = temp
    
    print(a)

基本上一旦有 a_bis = a 就表示上一个循环没有排序，所以 a 必须排序后才可以退出

Answer 3

您必须跟踪a[i] > a[i+1] 是否为真。如果没有，您的列表已排序，您可以跳出循环。

a = [27,21,22,1,11,23,0]
n = len(a)
while True:
    stop = True
    for i in range(n-1):
        if a[i]>a[i+1]:
            stop = False
            # a[i], a[i+1] = a[i+1], a[i]
            temp = a[i]
            a[i] = a[i+1]
            a[i+1] = temp
    if stop:
        break