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.

方法转自http://www.jianshu.com/p/bb84b0f866c9

问题关键在于判断两个字符串是否含有相同的字符。这里有一个好方法：因为我们不关心字符串是什么，只关心字符串中含有哪些字符，于是可以将每个字符串用一个int类型表示，a~z每个字母使用一位来存储（该位为1表示有这个字母，该位为0表示没有这个字母）。两个字符串没有相同字母则按位与的结果为0.

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