318. Maximum Product of Word Lengths

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本文介绍了一种算法，用于解决给定字符串数组中寻找两个不包含相同字母的单词，以获得它们长度的最大乘积。通过使用位运算，该算法能够高效地检查两单词是否包含重复字符。

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:

Input: ["abcw","baz","foo","bar","xtfn","abcdef"]
Output: 16
Explanation: The two words can be "abcw", "xtfn".
Example 2:

Input: ["a","ab","abc","d","cd","bcd","abcd"]
Output: 4
Explanation: The two words can be "ab", "cd".
Example 3:

Input: ["a","aa","aaa","aaaa"]
Output: 0
Explanation: No such pair of words.

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/maximum-product-of-word-lengths
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

判断是否有不同数字时，可以采用bit来记录每一个位。再取&，如果有相同的，那么&的结果肯定不是0；

class Solution {
    public int maxProduct(String[] words) {
        int ans = 0;
        for (int i = 0; i < words.length; i++) {
            for (int j = i + 1; j < words.length; j++) {
                if (isDiff(words[i], words[j])) {
                    ans = Math.max(ans, words[i].length() * words[j].length() );
                }
            }
        }
        return ans;
    }
    public boolean isDiff(String str1, String str2){
        int num1 = 0;
        int num2 = 0;
        for (char a: str1.toCharArray()) {
            num1 = num1 | 1 << (a - '0');
        }
        for (char a: str2.toCharArray()) {
            num2 = num2 | 1 << (a - '0');
        }
        int and = num1 & num2;
        return and == 0;
    }
}