題目描述：
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.
分析：
題目讓求的是兩個沒有任何相同字母的字符串長度相乘的最大值。本題的關鍵在於如何快速判斷兩個字符串是否存在相同字符。由於題目提到了所有的測試用例都只包含小寫字母，因此可以用一個int（32bit）前24bit分別表示a-z的24個字符是否出現。0表示沒有出現，1表示出現。然後字符串兩兩做"與"運算，結果爲0表示互相不包含相同字符串。再採用"打擂臺法"得出最大值即可。
代碼：
class Solution {
public int maxProduct(String[] words) {
// 核心：如何判斷兩個word"交叉"，採用位運算。
int[] wordBit = new int[words.length];
for(int i = 0;i<words.length;i++){
String word = words[i];
for(int j= 0 ;j<word.length();j++){
wordBit[i] |= 1<<(word.charAt(j) -'a');
}
}
int max = 0;
for(int i =0;i<wordBit.length;i++){
for(int j=i+1;j<wordBit.length;j++){
if((wordBit[i] & wordBit[j]) == 0){
// 無"交叉"
int l = words[i].length() *words[j].length();
max = max > l ? max : l;
}
}
}
return max;
}
}