Question

我无法使用以A，B，C开头的字符串对字符串的链接列表进行排序。因此，它只是一个链接列表，其中以'A'开头的所有字符串都出现在以'B'开头的所有字符串之前，所有这些都出现在以'C'开头的所有字符串之前。列表不需要进行排序，并且不需要保留以相同字母开头的字符串的相对顺序。它也需要在O（N）时间内。

我想做的方法是创建一个空的链表，然后通过给定的链表查找以A开头的所有字符串，然后将其添加到空列表中。然后再次查看给定列表中以B开头的字符串，然后将其添加到空列表中。我不确定这是否是O（N）时间。

Answer 1

This solution would require three O(N) traversings of the list, so it will still be O(N). The problem with it is that it creates a new list, and the requirements seem to imply inplace sorting.

An inplace approach could be to go over the list, and move any items starting with A to the beginning and any items starting with C to the end. E.g.:

for item in list:
    if item.data[0] == 'A'
        item.remove()
        list.prepend(item.data)
    elif item.data[0] == 'C'
        item.remove()
        list.append(item.data)

Notes:

item in this pseudocode snippet is the node object of the list - item.data is the data contained in it.
This solution assumes your list has prepend and append methods. If it doesn't, though, it shouldn't be too difficult to implement them.

Answer 2

You're close. Start with 3 empty lists, and distribute strings into those in a single pass through the original list. Keep a pointer to the last element of the 'A and 'B' lists (the first elements inserted) so that catenating the lists together doesn't require retraversal to find their ends.

按首字母排序字符串的链表

2 个答案: