Given a string array words, find the maximum value of length(word[i]) * length(word[j]) where the two words do not share common letters. You may assume that each word will contain only lower case letters. If no such two words exist, return 0.
Example 1:
Given ["abcw", "baz", "foo", "bar", "xtfn", "abcdef"]
Return 16
The two words can be "abcw", "xtfn".
Example 2:
Given ["a", "ab", "abc", "d", "cd", "bcd", "abcd"]
Return 4
The two words can be "ab", "cd".
Example 3:
Given ["a", "aa", "aaa", "aaaa"]
Return 0
No such pair of words.
由于只有26个小写字母，所以，我们可以为数组中的每个word开辟一个长为26的布尔数组作为哈希表，然后用一个两重for循环，两两比较，如果不存在公共的字母，则计算二者的长度的乘积，取最大作为最终结果。时间复杂度O(26n^2)，空间复杂度O(26n)。
上面的方法可以进一步优化，即长度为26的布尔数组，小于32位，可以编码为一个整数，这样两个整数按位与，如果结果为1，说明存在公共字母，如果结果为0，说明不存在公共字母。时间复杂度O(n^2)，空间复杂度O(n)。
{% if book.java %} {% codesnippet "./code/maximum-product-of-word-lengths-1."+book.suffix, language=book.suffix %}{% endcodesnippet %} {% endif %}
{% if book.java %} {% codesnippet "./code/maximum-product-of-word-lengths-2."+book.suffix, language=book.suffix %}{% endcodesnippet %} {% endif %}