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 stub
Set<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;
}
}