Python 我在字谜中对单词进行分组的时间超过了，还有其他方法吗

上述代码在馈送时为：

class Solution:
# @param {string[]} strs
# @return {string[][]}
def isAnagram(self, s, t):
    s1 = [];s2 = []
    s1[:0] = s
    s2[:0] = t
    s1.sort();s2.sort()
    if s1 == s2:
        return True
    else:
        return False
def groupAnagrams(self, strs):
    l=[]
    for i in strs:
        m=[]
        if m not in m:
            m.append(i)
            for j in strs:
                if self.isAnagram(i,j)==True:
                    m.append(j)
                    strs.remove(j)
        l.append(m)
    return l

显示了这个错误。

---

我是python新手，所以我无法跨越这一关。谢谢：）

您的代码似乎有一些问题

• 如果m不在m中：

  这行没有多大意义

• strs.remove（j）

  在迭代列表时永远不要从列表中删除，否则会发生不好的事情
• 您正在将每个字符串与其他字符串（包括字符串本身）进行比较

因此，对于您的

[“吃”、“茶”、“晒”、“吃”、“纳特”、“蝙蝠”]

示例，您的代码返回

[[“吃”、“吃”、“吃”]、[“晒”、“晒”]、[“蝙蝠”、“蝙蝠”]

关于性能，最大的问题似乎是，每次将字符串与任何其他字符串进行比较时，您都在对这些字符串进行排序，即，总的来说，您排序了n2次！相反，我建议您使用字典将字符串映射到它们的排序版本

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arrange","zippy","flosses","parnell","erratas","sidings","clapboards","confederated","palliative","wirelesses","etruscans","neonates","clayey","vaccinating","peskiest","liable","bibliographical","squidded","hausdorff","lumberyard","blythe","pillions","fiddlesticks","sarong","scarfed","reformer","gunrunning","sweaters","entreats","wicca","tennis","quilt","canisters","frankincense","unbar","neighed","cicadas","bighorns","tittles","dimaggio","costuming","judas","paints","pastorals","carib","glamored","cantering","demotes","currying","excommunicating","thwarting","freebase","niagara","fortification","buttercups","survey","barracudas"]

---

示例的输出是

[['tan'，'nat']，['bat']，['eat'，'tea'，'ate']]

您可以这样做。在这里，我将每个单词进行散列，并绘制单词地图。具有相同哈希值的单词是字谜

def groupAnagrams(strs):
    res = {}
    for s in strs:
        res.setdefault(''.join(sorted(s)), []).append(s)
    return res.values()

---

我假设您正在某种托管/教育环境中运行代码？你能给我们一点关于你想做什么的背景知识吗（一大堆句子？），并正确地格式化你的代码（类缩进看起来不合适）？例如，给定：[“吃”、“茶”、“晒”、“吃”、“吃”、“吃”、“茶”]，[“吃”、“晒”]，[“拍”]]我们需要一个输出：[“吃”、“吃”、“茶”]，[“nat”、“晒”]，[“拍”]你的代码有问题吗（就耗时而言）是将每个字符串与其他字符串进行比较。因此，如果有1000个字符串，它将进行1000*1000（=1000000）比较。这称为“N平方阶”或“O（N^2）”，这意味着它将非常快地变得非常慢。您需要更改算法，以便不需要所有这些比较。提示：也许可以考虑将每个字符串预处理为一个“否”，但不会这样做，一旦它找到其字谜，我会将其从原始列表中删除，但如果大多数单词不是彼此的字谜呢？

import collections


def sort_prehash(word):
    return ''.join(sorted(word))  

def group_anagrams(words, hash_function):
    result = {}
    for w in words:
        s = hash_function(w.lower())
        if s in result:
            result[s] |= {w}
        else:
            result[s] = {w}
    return result.values()

orig = ["eat", "tea", "tan", "ate", "nat", "bat"]

print group_anagrams(orig, sort_prehash)