LeetCode Medium: 49. Group Anagrams

一、题目

Given an array of strings, group anagrams together.

Example:

Input: ["eat", "tea", "tan", "ate", "nat", "bat"],
Output:
[
  ["ate","eat","tea"],
  ["nat","tan"],
  ["bat"]
]

Note:

• All inputs will be in lowercase.
• The order of your output does not matter.

---

给定一组字符串，相同的分组

二、思路

采用字典，首先采用类似于one-hot编码，将26个字母作为基体，对于每个字符串，相应位置的字母做计数操作，如果字符串字母组成相同但是顺序不同，但是编码之后的形式肯定是相同，用编码后的形式作为key，对应的字符串作为value。

 三、代码

#coding:utf-8
import collections
class Solution:
    def groupAnagrams(self, strs):
        """
        :type strs: List[str]
        :rtype: List[List[str]]
        """
        resdic = collections.defaultdict(list)  #这句代码很重要，在Python中如果访问字典中不存在的键，会引发KeyError异常。因此，可以采用collections.defaultdict来初始化字典。defaudict初始化函数接受一个类型作为参数，当访问不存在的key时，可以实例化一个值作为默认值，从而省去了使用dict时需要判断key是否存在的问题。试过，如果不这样做的话会报错
        dic = {'a':0,'b':1,'c':2,'d':3,'e':4,'f':5,'g':6,'h':7,'i':8,'j':9,'k':10,'l':11,'m':12,'n':13,'o':14,'p':15,'q':16,'r':17,'s':18,'t':19,'u':20,'v':21,'w':22,'x':23,'y':24,'z':25}
        for str in strs:
            count = [0]*26
            for c in str:
                count[dic[c]] += 1
            resdic[tuple(count)].append(str)
        print('me:',list(resdic.values()))
        return list(resdic.values())

    def groupAnagrams1(self,strs):
        ans = collections.defaultdict(list)
        for s in strs:
            count = [0] * 26
            for c in s:
                count[ord(c) - ord('a')] += 1
            ans[tuple(count)].append(s)
        print('not me:',list(ans.values()))
        return list(ans.values())
if __name__ == '__main__':
    ss = Solution()
    a = ["eat","tea","tan","ate","nat","bat"]
    ss.groupAnagrams(a)
    ss.groupAnagrams1(a)