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.

/* * 用个数组表示每个string，通过比较每个数组有没有重叠来判断有没有相同的字符 * 转念一想，用什么数组，一个integer足矣 */class Solution {    public int maxProduct(String[] words) {        int[] a = new int[words.length];        for(int i=0; i<words.length; i++) {        for(char c : words[i].toCharArray())        a[i] |= (1<<(c-'a'));        }                int max = 0;        for(int i=0; i<a.length; i++)        for(int j=i+1; j<a.length; j++)        if((a[i] & a[j]) == 0)        max = Math.max(max, words[i].length() * words[j].length());                return max;    }}