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.

这个题用蛮力法超时：

public class Solution {    public int maxProduct(String[] words) {        int n=words.length;        int max=0;        for(int i=0;i<n-1;i++){        for(int j=i+1;j<n;j++){        if(!haveCommonLetter(words[i],words[j])){        int product=words[i].length()*words[j].length();        max=product>max?product:max;        }        }        }        return max;    }private boolean haveCommonLetter(String word1, String word2) {// TODO Auto-generated method stubSet<Character> set=new HashSet<Character>();for(int i=0;i<word1.length();i++){if(!set.contains(word1.charAt(i))){set.add(word1.charAt(i));}}for(int i=0;i<word2.length();i++){if(set.contains(word2.charAt(i))){return true;}}return false;}}

正确解法如下：

用一个int表示一个word的情况，每一个bit记录字母是否存在，然后&操作符就可以相当于求交集，实在是巧妙，值得学习。

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