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.

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 1 public class Solution {
 2     public int maxProduct(String[] words) {
 3         int ans = 0;
 4         int[] bitmap = new int[words.length];
 5         
 6         for(int i = 0; i < words.length; i++){
 7             for(int j = 0; j < words[i].length(); j++){
 8                 bitmap[i] |= 1 << (words[i].charAt(j) - 'a');
 9             }
10         }
11         
12         for(int i = 0; i < bitmap.length; i++){
13             for(int j = i+1; j < bitmap.length; j++){
14                 if((bitmap[i] & bitmap[j]) == 0 && words[i].length()*words[j].length() > ans){
15                     ans = words[i].length()*words[j].length();
16                 }
17             }
18         }
19         
20         return ans;
21     }
22 }

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。