Maximum Product of Word Lengths

描述

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.

分析

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)

解法1

解法2

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