318. Maximum Product of Word Lengths

我一開始只想到到最差的作法, 哇用bit這招真的高明, 然後只要兩數互相and, 不為0就代表一定有重複的字母

時間複雜度為O(n²)

class Solution {
public:
    int maxProduct(vector<string>& words) {
        vector<int> mask(words.size());
        int res = 0;
        for(int i = 0; i < words.size();i++){
            for(char c: words[i]){
                mask[i] |= 1 << (c - 'a');
            }
            for(int j = 0; j < i;j++){
                if((mask[i] & mask[j]) == 0){
                    res = max(res, (int)(words[i].size() * words[j].size()));
                }
            }
        }
        return res;
    }
};