[leetcode] 318. 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.

解法一：

一开始以为有linear的解法，没有想出来。后来O(n^2)的解法用hash table还是不行。

现在这个思路是把每一个单词转成一个int(32位)，出现该字母就将对应的bit置一。如果两个单词具有相同的字母，那么他们的&操作不为0。

class Solution {
public:
int maxProduct(vector<string>& words) {
int res = 0;
int num = words.size();
vector<int> m(num,0);
for(int i=0; i<num;++i){
for(char a:words[i])
m[i] |= 1<<(a-'a');

for(int j=0;j<i;++j){
if((m[i]&m[j])==0){
res = max(res,int(words[i].size()*words[j].size()));
}
}
}
return res;

}
};