题目描述：
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.
解题思路：
看到讨论区里面别人的解法，实在是太赞了。
思想是对于每个字符串，统计他所对应的字母出现与否，这是通过移位运算符来实现的。具体的，对于出现a，则将最右边的位数置为1，如果出现b，则将1向右移动一位，将第二位的数字置为1；然后通过或运算实现对应位置的表示，如出现了a，则最右边的位置为1；出现了c，则从右向左数第三位的数字为1；
然后将每个字符串互相比较，通过位与运算符来比较，如果两个字符串没有重叠的字母，那么位与之后应该结果为0，否则为1；
然后判断位数相乘的结果
代码如下：
public class Solution{
public int maxProduct(String[] words){
if(words == null || words.length == 0)
return 0;
int len = words.length;
int[] num = new int[len];
int maxProduct = 0;
for(int i = 0; i < len; i++){
String temp = words[i];
for(int j = 0; j < temp.length(); j++){
num[i] |= (1 << (temp.charAt(j) - ‘a‘));
}
}
for(int i = 0; i < len; i++){
for(int j = i + 1; j < len; j++){
if((num[i] & num[j]) == 0){
int temp = words[i].length() * words[j].length();
if(temp > maxProduct)
maxProduct = temp;
}
}
}
return maxProduct;
}
}
Java [Leetcode 318]Maximum Product of Word Lengths
原文：http://www.cnblogs.com/zihaowang/p/5315828.html