给你一个字符串数组 words ，找出并返回 length(words[i]) * length(words[j]) 的最大值，并且这两个单词不含有公共字母。如果不存在这样的两个单词，返回 0 。

示例 1：

输入：words = ["abcw","baz","foo","bar","xtfn","abcdef"]
输出：16 
解释：这两个单词为 "abcw", "xtfn"。

var maxProduct = function(words) {
    const length = words.length;
    const masks = new Array(length).fill(0);
    for (let i = 0; i < length; i++) {
        const word = words[i];
        const wordLength = word.length;
        for (let j = 0; j < wordLength; j++) {
            masks[i] |= 1 << (word[j].charCodeAt() - 'a'.charCodeAt());
        }
    }
    let maxProd = 0;
    for (let i = 0; i < length; i++) {
        for (let j = i + 1; j < length; j++) {
            if ((masks[i] & masks[j]) === 0) {
                maxProd = Math.max(maxProd, words[i].length * words[j].length);
            }
        }
    }
    return maxProd;
};

来源：力扣（LeetCode） 链接：https://leetcode-cn.com/problems/maximum-product-of-word-lengths