318. Maximum Product of Word Len

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.

一刷
题解：
首先每个string用一个int变量，如果为二进制，如果第i位为1，表示字符串中有'a' + i这个字母。
然后进行O(n^2)的搜索，只有当两个string的变量与之后仍为0（没有相同的字母），计算长度积，并且更新res

public static int maxProduct(String[] words) {
    if (words == null || words.length == 0)
        return 0;
    int len = words.length;
    int[] value = new int[len];
    for (int i = 0; i < len; i++) {
        String tmp = words[i];
        value[i] = 0;
        for (int j = 0; j < tmp.length(); j++) {
            value[i] |= 1 << (tmp.charAt(j) - 'a');
        }
    }
    int maxProduct = 0;
    for (int i = 0; i < len; i++)
        for (int j = i + 1; j < len; j++) {
            if ((value[i] & value[j]) == 0 && (words[i].length() * words[j].length() > maxProduct))
                maxProduct = words[i].length() * words[j].length();
        }
    return maxProduct;