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:
Input: ["abcw","baz","foo","bar","xtfn","abcdef"]
Output: 16
Explanation: The two words can be "abcw", "xtfn".
Example 2:
Input: ["a","ab","abc","d","cd","bcd","abcd"]
Output: 4
Explanation: The two words can be "ab", "cd".
Example 3:
Input: ["a","aa","aaa","aaaa"]
Output: 0
Explanation: No such pair of words.
public static int maxProduct(String[] words) {
if (words == null || words.length == 0)
return 0;
int len = words.length;
int[] value = new int[len];
for (int i = 0; i < len; i++) {
String tmp = words[i];
value[i] = 0;
for (int j = 0; j < tmp.length(); j++) {
value[i] |= 1 << (tmp.charAt(j) - 'a');
}
}
int maxProduct = 0;
for (int i = 0; i < len; i++)
for (int j = i + 1; j < len; j++) {
if ((value[i] & value[j]) == 0 && (words[i].length() * words[j].length() > maxProduct))
maxProduct = words[i].length() * words[j].length();
}
return maxProduct;
}
注意：可以通过掩码来判断两个string是否包含重复的字符！！！！
