#yyds干货盘点# LeetCode程序员面试金典：最大单词长度乘积

题目

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"。

示例 2：

输入：words = ["a","ab","abc","d","cd","bcd","abcd"]

输出：4  

解释：这两个单词为 "ab", "cd"。

示例 3：

输入：words = ["a","aa","aaa","aaaa"]

输出：0  

解释：不存在这样的两个单词。

代码实现

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