Maximum Product of Word Lengths(Medium)

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.

解题思路： 因为每个字母都是小写字母，所以我们可以用一个int的26位去保存每个单词所包含的字母的信息：因为我们不关心每个单词中各个字母出现的个数，只关心是否出现，所以可以用1表示出现，0表示未出现。这样的好处是，经过这样的预处理之后，当需要判断两个单词是否有相同字母时，做个与计算就知道结果了（是否为0）。这样，我们就得到了O(n*n)的解法。

public int maxProduct(String[] words) {
    int[] wordMask = new int[words.length];

    for (int i = 0; i < words.length; i++) {
        wordMask[i] = 0;
        String word = words[i];
        for (char c : word.toCharArray()) {
            wordMask[i] =  wordMask[i] | (1 << (c - 'a'));
        }
    }

    int max = 0;
    for (int i = 0; i < wordMask.length - 1; i++) {
        for (int j = i+1; j < wordMask.length; j++) {
            if ((wordMask[i] & wordMask[j]) == 0) {
                int tmp = words[i].length() * words[j].length();
                if (max < tmp) {
                    max = tmp;
                }
            }
        }
    }

    return max;
}