318. Maximum Product of Word Lengths 难度：easy

题目：

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.

思路：

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。

程序：

class Solution {
public:
int maxProduct(vector<string>& words) {

int len = words.size();

vector<int> num(len,0);
int res = 0;

for(int i = 0;i < len;i++)
for(int j = 0;j < words[i].size();j++)
num[i] |= (1 << (words[i][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].size() * words[j].size();
if(temp > res)
res = temp;
}
}
}
return res;
}
};