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 static int maxProduct(String[] words){
        int[] arri = new int[words.length];
        for (int i = 0; i < words.length; i++) {
            for (int j = 0; j < words[i].length(); j++) {
                arri[i] |= 1<<(words[i].charAt(j)-'a');
            }
        }

        int ans = 0;
        for (int i = 0; i < words.length-1; i++) {
            for (int j = i+1; j < words.length; j++) {
                if ((arri[i] & arri[j]) == 0) {// 没有重复的字母
                    ans = Math.max(ans, words[i].length()*words[j].length());
                }
            }
        }
        return ans;
    }

移位运算符：

1 << 0 => 1

1 << 1 => 2

1 << 2 => 4

1 << 3 => 8

1 << 4 => 16

1 << 5 => 32

1 << 6 => 64

1 << 7 => 128

1 << 8 => 256

1 << 9 => 512

1 << 10 => 1024

1 << 11 => 2048

1 << 12 => 4096

1 << 13 => 8192

1 << 14 => 16384

1 << 15 => 32768

1 << 16 => 65536

1 << 17 => 131072

1 << 18 => 262144

1 << 19 => 524288

1 << 20 => 1048576

1 << 21 => 2097152

1 << 22 => 4194304

1 << 23 => 8388608

1 << 24 => 16777216

1 << 25 => 33554432

 

words[i] 对应着arri[i]:words[i]中的每个单词编码后得到arri[i]

注释处相交等于0说明有重复字母。