这具有O(n)的优势(将字符串排序为O(n log n)).如果Python中的一个字符(或字符串)以字母顺序出现在另一个字符之前，则它小于"另一个字符，因此，为了查看字符串是否以字母顺序排列，我们只需要比较每对相邻的字符即可.另外，请注意，您使用range(len(word)-1)而不是range(len(word))，因为否则，在循环的最后一次迭代中，您将越过字符串的边界.

This has the advantage of being O(n) (sorting a string is O(n log n)). A character (or string) in Python is "less than" another character if it comes before it in alphabetical order, so in order to see if a string is in alphabetical order we just need to compare each pair of adjacent characters. Also, note that you take range(len(word) - 1) instead of range(len(word)) because otherwise you will overstep the bounds of the string on the last iteration of the loop.

def isInAlphabeticalOrder(word):
    for i in range(len(word) - 1):
        if word[i] > word[i + 1]:
            return False
    return True