318. 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.

https://leetcode.com/discuss/74580/bit-shorter-c

判断两个字符串有没有共同的字符（都是小写字母）

class Solution {
public:
int maxProduct(vector<string>& words) {
int result = 0;
if(words.empty()) return result;
unordered_map<int ,int> maps;
for(auto word : words){
int mask = 0;
for(auto c : word)
mask |= 1<<(c-'a');   //这句是精华，用比特位指示出现某个字符
maps[mask] = max(maps[mask],(int) word.size());
}
for(auto a : maps)
for(auto b : maps){
if(!(a.first & b.first))
result = max(result, a.second*b.second);
}
return result;
}
};