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Given an array of strings strs, group the anagrams together. You can return the answer in any order.
An Anagram is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.
Example 1:
<pre>Input: strs = ["eat","tea","tan","ate","nat","bat"] Output: [["bat"],["nat","tan"],["ate","eat","tea"]] </pre>
Example 2:
<pre>Input: strs = [""] Output: [[""]] </pre>
Example 3:
<pre>Input: strs = ["a"] Output: [["a"]] </pre>
Constraints:
1 <= strs.length <= 10<sup>4</sup>0 <= strs[i].length <= 100strs[i]consists of lowercase English letters.
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题目大意：
对同字母不同序单词分组
算法思路：
N/A
注意事项：
- list(id_to_words.values())要转成list
Python代码：
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13def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
id_to_words = collections.defaultdict(list)
for word in strs:
id_to_words[self.get_id(word)].append(word)
return list(id_to_words.values()) # remember to convert it to list
def get_id(self, word):
char_to_freq = collections.Counter(word)
res = ''
for c in string.ascii_lowercase:
if c in char_to_freq:
res += c + str(char_to_freq[c])
return res
算法分析：
时间复杂度为O(nm)，空间复杂度O(n+m). n是单词个数，m是单词长度
算法II解题思路(推荐)：
用排序作为id
注意事项：
- sorted(word)返回的是排好序的字母列表，要用join转回单词
- list(anagrams_dict.values())要转成list
Python代码：
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6def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
anagrams_dict = collections.defaultdict(list)
for word in strs:
_id = "".join(sorted(word))
anagrams_dict[_id].append(word)
return list(anagrams_dict.values())
