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.

给定一个字符串数组，从中找出两个字符串使它们的乘积最大，但是这两个字符串中不存在相同的字符。如果直接查找，数组中每两个字符串都要相互比较，字符串中的字符也要比较，时间复杂度太大。我们可以采用位运算，先将字符串中出现的字符用二进制位来表示，一共有26个字符，我们用一个32位的整数就可以表示。然后判断两个字符串中是否存在相同的字符时，可以将两个字符串的对应的二进制数进行位与运算，如果为0说明不存在相同字符，如果不为0说明存在相同字符。时间复杂度为O(n^2)。代码如下：

public class Solution {
    public int maxProduct(String[] words) {
        if(words == null || words.length == 0) return 0;
        int max = 0;
        int[] bitString = new int[words.length];
        for(int i = 0; i < words.length; i++) {
            char[] w = words[i].toCharArray();
            for(int j = 0; j < w.length; j++) {
                bitString[i] |= 1 << (w[j] - 'a');
            }
        }
        
        for(int i = 0; i < bitString.length - 1; i++) {
            for(int j = i + 1; j < bitString.length; j++) {
                if((bitString[i] & bitString[j]) == 0)
                    max = Math.max(max, words[i].length() * words[j].length());
            }
        }
        return max;
    }
}