leetcode 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.

题意：求，不包含相同字符的两个string的最大长度积；

思路：

如何降低判断两string是否包含相同字符的开销？？？

用bit来表示string元素中a-z每个字符是否出现；

class Solution {
public:
    int maxProduct(vector<string>& words) {
        int len = words.size();
        if(len==0)
            return 0;
        vector<int> v(len,0);
        vector<int> l(len,0);
        for(int i=0;i<len;i++)
        {
            string tstr = words[i];
            int tlen = tstr.length();
            l[i] = tlen;
            for(int j=0;j<tlen;j++)
            {
                int ttmp = 1<<(tstr[j]-'a');
                v[i] |= ttmp;
            }
        }

        int ans = 0;
        for(int i=0;i<len;i++)
        {
            for(int j=i+1;j<len;j++)
            {
                int t2 = v[i]&v[j];
                if(t2!=0)
                    continue;
                int tmp = l[i]*l[j];
                if(tmp>ans)
                    ans = tmp;
            }
        }
        return ans;
    }
};