49-Group Anagrams-(Medium) 题解

1、题目

　　

Given an array of strings, group anagrams together.

For example, given: ["eat", "tea", "tan", "ate", "nat", "bat"], 
Return:

[
  ["ate", "eat","tea"],
  ["nat","tan"],
  ["bat"]
]
Note: All inputs will be in lower-case.

2、分析，该题目可以针对每个字符串进行排序，然后计算每个字符串的hash值进行归类

3、优化、在计算hash值的时候，考虑hash算法跟字符串排序无关，也就是不用进行排序，针对26个字母分配26个质数，然后将hash函数设置为每个字母映射的质数乘积，便不用进行排序了，该方法利用了质数的特性

　 两个质数相乘得到一个合数，这个合数不会分解为其它质数的乘积

4、代码：

　　

#!/usr/local/bin/python3
# -*- coding: utf-8 -*-
__author__ = 'qqvipfunction'


primeTable = [2,3,5,7,11, 13,17,19,23,29, 31,37,41,43,47, 53,59,61,67,71, 73,79,83,89,97, 101]

class Solution(object):

    def groupAnagrams(self, strs):
        """
        :type strs: List[str]
        :rtype: List[List[str]]
        """
        map = {}
        for i in range(0 , len(strs)):
            str = strs[i]
            hash = self.hash_str(str)
            list = map.get(hash, None)
            if not list:
                list = []
            list.append(str)
            map[hash] = list

        return map.values()

    def hash_str(self, str):
        length = len(str)
        charAvalue = ord('a')
        if length > 0:
            hashSum = 1
            for i in range(0, length):
                hashSum = hashSum * primeTable[(ord(str[i]) - charAvalue)]
            return hashSum
        return 0



if __name__ == '__main__':
    s = Solution()
    print(s.groupAnagrams(["eat", "tea", "bat"]))