Leetcode: Maximum Product of Word Lengths

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.

思路：

本题关键在于需要判断两个字符串是否交叉，如果使用string的find功能，那效率会比较低，可能无法接受。换一种思路，由于限定了所有字符串都是小写字母，所以我们可以用位操作来实现，每个字符对应int中的一个位，如果出现则这个位置记为1，否则为0，最终每个字符串都会得到一个int值，标记为Key，那判断两个字符串是否交叉就只需判断两个Key按位与是否为0，为0则不交叉，否则存在交叉。代码实现如下：

class Solution {public:    static int maxProduct(vector<string>& words) {        vector<int> wordsKey(words.size(), 0);        int value = 0;        for (size_t i = 0; i < words.size(); ++i)        {            string &curword = words[i];            for (size_t j = 0; j < curword.size(); ++j)                wordsKey[i] |=  0x1 << (curword[j] - 'a');        }        for (size_t i = 1; i < wordsKey.size(); ++i)        {            for (size_t j = 0; j < i; j++)            {                if ((wordsKey[i] & wordsKey[j]) == 0 && words[i].size() * words[j].size() > value)                    value = words[i].size() * words[j].size();            }        }        return value;    }};