Maximum Product of Word Lengths.md

#Maximum Product of Word Lengths ##Problem: 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. ##Idea: 暴力求解method of exhaustion，用bool[26]记录字母是否出现过。 ##To Study: 可以用mask(int型)来记录每个字母是否出现
用 & 判断是否有共同的字母

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