[LeetCode]318. Maximum Product of Word Lengths-优快云博客

本文探讨了如何在给定字符串数组中寻找两个不含有共同字母的单词，并计算这两个单词长度的乘积的最大值。通过使用位运算进行优化，提供了一个高效算法实现。

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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.

35ms:

     public int maxProduct(String[] words) {
        int max = 0;
        int[] bytes = new int[words.length];
        for(int i=0;i<words.length;i++){
            int val = 0;
            for(int j=0;j<words[i].length();j++){
                val |= 1<<(words[i].charAt(j)-'a');
            }
            bytes[i] = val;
        }
        for(int i=0; i<bytes.length; i++){
            for(int j=i+1; j<bytes.length; j++){
                if((bytes[i] & bytes[j])==0)max = Math.max(max,words[i].length()*words[j].length());
            }
        }
        return max; 
     }