318. Maximum Product of Word Lengths java solutions

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.

Credits:
Special thanks to @dietpepsi for adding this problem and creating all test cases.

Subscribe to see which companies asked this question

 

public class Solution {
    public int maxProduct(String[] words) {
        int ans = 0;
        int[] bitmap = new int[words.length];
        
        for(int i = 0; i < words.length; i++){
            for(int j = 0; j < words[i].length(); j++){
                bitmap[i] |= 1 << (words[i].charAt(j) - 'a');
            }
        }
        
        for(int i = 0; i < bitmap.length; i++){
            for(int j = i+1; j < bitmap.length; j++){
                if((bitmap[i] & bitmap[j]) == 0 && words[i].length()*words[j].length() > ans){
                    ans = words[i].length()*words[j].length();
                }
            }
        }
        
        return ans;
    }
}

之前都没有做过类似的题目，题目要找出字符串数组中两个 没有一个字母相同的字符串乘积的最大值，学习了大神的做法。

字母只有26个，java 中一个int 有32位。

通过 bitmap[i] |= 1 << (words[i].charAt(j) - 'a'); 即可记录下下标为i的字符串字符特征。

全部扫描一遍，即可记录所有的字符串特征， 之后二重循环即可得答案。