LeetCode刷题【318】最大单词长度乘积

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。

提示：

• 2 <= words.length <= 1000
• 1 <= words[i].length <= 1000
• words[i] 仅包含小写字母

位运算->哈希，位运算->重复判断

抄大佬作业，这个我是真想不来

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