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.

    用位操作，用一个int代表26个字母， 
    如 0000 0000 0000 0001 表示 a， 
    用 0000 0000 0000 0101 代表 ac .  
    判断两个字符串有没有相同字母，就用 两个int 相与 （&）， 若为 0 则表示两字符串没有相同字母。
    代码如下：

class Solution {
public:
    int maxProduct(vector<string>& words) {
        if (words.size() <= 1) return 0;
        int * bit = new int[words.size()];

        for (int i = 0; i < words.size(); i++) {
            bit[i] = 0;
            for (int j = 0; j < words[i].size(); j++) {
                bit[i] |= 1 << (words[i][j] - 'a');
                // a is 0000 0000 0000 0001
            }
        }

        int ret = 0;
        for (int i = 0; i < words.size() - 1; i++) {
            for (int j = i + 1; j < words.size(); j++) {
                if (0 == (bit[i] & bit[j])) {
                    int temp = words[i].size() * words[j].size();
                    ret = (ret < temp ? temp : ret);
                }
            }
        }

        delete [] bit;
        return ret;
    }
};