318. 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.

public class Solution {    public int maxProduct(String[] words) {        int res = 0;if (words.length <= 1)return 0;int[] numbers = new int[words.length];// 把字符串用二进制保存for (int i = 0; i < words.length; i++) {for (int j = 0; j < words[i].length(); j++) {numbers[i] |= 1 << (words[i].charAt(j) - 'a');}}for (int i = 0; i < words.length - 1; i++) {for (int j = i + 1; j < words.length; j++) {if ((numbers[i] & numbers[j]) == 0) {res = Math.max(res, words[i].length() * words[j].length());}}}return res;    }}