1）先对所有字符串重排序，然后取set获得总共的分组个数。再逐个判断加入。时间复杂度太高了
class Solution: def groupAnagrams(self, strs: List[str]) -> List[List[str]]: def so(x): x= sorted(x) return ‘‘.join(x) temp = list(map(so, strs)) set_temp = list(set(temp)) result = list() for i in range(len(set_temp)): result.append([]) for i in range(len(temp)): result[set_temp.index(temp[i])].append(strs[i]) return result
2）哈希表+排序！字典的values也可以是列表！空间复杂度很低
class Solution: def groupAnagrams(self, strs: List[str]) -> List[List[str]]: dic = dict() for i in strs: temp = ‘‘.join(sorted(i)) if temp not in dic.keys(): dic[temp] = [i] else: dic[temp].append(i) return list(dic.values())
原文：https://www.cnblogs.com/cbachen/p/14881477.html