1. navie：集合 intersect

以集合的形式分别存放两字符串，然后求集合交集。

def common_words_naive(str1, str2):
	str1_set = set(str1.strip().split())
	str2_set = set(str2.strip().split())
	return str1_set & str2_set			# 集合 intersect

>> common_words_naive('I love word', 'I love China')
{'I', 'love'}

2. 使用 hash

• 根据字符串hash算法，对字符串1的单词分别求其hash值，时间空间复杂度均为 O ( n ) O(n) O(n)，并将hash值，存放在集合中

• 遍历字符串2中的单词，求其hash值，判断是否在字符串1的hash集合中，如果是，则为 common words

  def bkdr_hash(str, seed=131):
      hash = 0
      for s in str:
          hash = hash*seed + ord(s)
      return hash & 0x7fffffff
  

  将字符串hash为整数值的方法及其对比见种字符串Hash函数比较

  def common_words_hash(str1, str2):
      words = str1.strip().split(' ')
      str1_hashset = set(bkdr_hash(word) for word in words)
      common_words = []
      for word in str2.strip().split(' '):
          if bkdr_hash(word) in str1_hashset:
              common_words.append(word)
      return common_words
      
  >> common_words_hash('I love word', 'I love China')
  {'I', 'love'}
  

references

https://www.glassdoor.com/Interview/Given-2-strings-find-the-common-words-along-with-the-time-and-space-complexity-How-would-you-optimize-the-algorithm-QTN_341530.htm#