318. 最大单词长度乘积.

 

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给你一个字符串数组 words ，找出并返回 length(words[i]) * length(words[j]) 的最大值，并且这两个单词不含有公共字母。如果不存在这样的两个单词，返回 0 。

 

示例 1：

["abcw","baz","foo","bar","xtfn","abcdef"]
16 
解释
这两个单词为 "abcw", "xtfn"

示例 2：

["a","ab","abc","d","cd","bcd","abcd"]
4 
解释
"ab", "cd"

示例 3：

["a","aa","aaa","aaaa"]
0 
解释
不存在这样的两个单词。

 

class Solution {
public:
    int maxProduct(vector<string>& words) {
        int n = words.size();
        vector<int>mask(n,0);
        for(int i = 0; i < n;i++) {
            for(char cc: words[i]) {
                mask[i] |= (1<<(cc-'a'));
            }
        }
        int res = 0;
        for(int i = 0; i < n;i++) {
            for(int j = i+1; j< n;j++) {
                if ((mask[i]& mask[j])==0) {
                    res = max(res, int(words[i].size())*int(words[j].size()));
                }
            }
        }
        return res;
    }
};