Question

我正在尝试解决Hackerrank问题：为给定的字符串输入找到下一个按字典顺序更大的字符串排列。

这是我的解决方法：

def biggerIsGreater(w):

    if len(w)<=1: return w

    # pair letters in w string with int representing positional index in alphabet
    letter_mapping = dict(zip(string.ascii_lowercase, range(1, len(string.ascii_lowercase)+1)))

    char_ints = [letter_mapping[letter] for letter in w.lower() if letter in letter_mapping]

    # reverse it
    reversed_char_ints = char_ints[::-1]

    # get char set to reorder, including pivot.
    scanned_char_ints = []
    index = 0
    zipped = list(zip(reversed_char_ints, reversed_char_ints[1:]))
    while index < len(zipped):
        char_tuple = zipped[index]
        scanned_char_ints.append(char_tuple[0])
        if char_tuple[0] <= char_tuple[1]:
            if index == len(zipped) - 1:
                return "no answer"
        else:
            scanned_char_ints.append(char_tuple[1])
            break
        index += 1

    # get smallest among bigger values of pivot
    char_to_switch = None
    char_to_switch_index = None
    for item in scanned_char_ints[:-1]:
        if item > scanned_char_ints[-1]:
            if char_to_switch == None or item <= char_to_switch:
                char_to_switch = item
                char_to_switch_index = scanned_char_ints.index(item)

    # switch pivot and smallest of bigger chars in scanned chars
    pivot_index = len(scanned_char_ints) - 1
    scanned_char_ints[pivot_index], scanned_char_ints[char_to_switch_index] = scanned_char_ints[char_to_switch_index], scanned_char_ints[pivot_index]

    # order from second to end the other chars, so to find closest bigger number of starting number
    ord_scanned_char_ints = scanned_char_ints[:-1]
    ord_scanned_char_ints.sort(reverse=True)
    ord_scanned_char_ints.append(scanned_char_ints[-1])

    # reverse scanned chars
    ord_scanned_char_ints.reverse()

    # rebuild char int list
    result_ints = char_ints[:len(char_ints) - len(ord_scanned_char_ints)]

    result_ints.extend(ord_scanned_char_ints)

    result_ = ""
    for char_intx in result_ints:
        for char, int_charz in letter_mapping.items():
            if int_charz == char_intx:
                result_ += char
    return result_

（我知道互联网上有解决方案，可以用更简洁的方式解决问题，但我显然很想自己成功）。现在，它似乎可以运行1、2、100个最多100个字符的字符串输入。但是，当hackerrank测试过程针对最多100个字母的100000个字符串进行测试时，将导致错误，并且没有关于它的更多信息。在我的机器上运行具有类似输入大小的测试不会引发任何错误。

此解决方案有什么问题？

预先感谢

字典顺序上的下一个更大的字符串排列和解决方案效率

0 个答案: