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.

Credits:
Special thanks to @dietpepsi for adding this problem and creating all test cases.

思路：用hashmap来统计，会超时，题目提示，只包含小写字母，这样只有26位，用位移来做，只要两者相与的结果为0，则为互斥关系，更新最大值即可。

public class Solution {
    public int maxProduct(String[] words) {
        if(words == null || words.length == 0) return 0;
        int[] mask = new int[words.length];
        for(int i=0; i<words.length; i++){
            for(int j=0; j<words[i].length(); j++){
                mask[i] |= 1<<(words[i].charAt(j) - 'a');
            }
        }
        
        int res = 0;
        for(int i=0; i<words.length-1; i++){
            for(int j=i+1; j<words.length; j++){
                if((mask[i] & mask[j])==0) {
                    res = Math.max(res, words[i].length() * words[j].length());
                }
            }
        }
        return res;
    }
}