剑指 Offer II 005. 单词长度的最大乘积

去重，然后暴力即可，去重时别忘了记录原来的长度

class Solution {
public:
    int max(int a, int b)
    {
        return a > b ? a : b;
    }
    int maxProduct(vector<string>& words) {
        int len = words.size();
        vector<string> ww;
        int a[1001];
        bool vis[26];
        bool vv[1001][26];
        memset(vv, 0, sizeof(vv));
        for(int i = 0; i < len; i++)
        {
            string temp = "";
            memset(vis, 0, sizeof(vis));
            a[i] = words[i].length();
            for(int j = 0; j < a[i]; j++)
                if(vis[words[i][j] - 'a'] == 0)
                {
                    temp += words[i][j], vis[words[i][j] - 'a'] = 1;
                    vv[i][words[i][j] - 'a'] = 1;
                }
            ww.push_back(temp);
        }
        int sum = 0;
        for(int i = 0; i < len; i++)
        {
            string temp = ww[i];
            int len1 = temp.length();
            bool flag = 0;
            for(int j = i + 1; j < len; j++)
            {
                flag = 0;
                for(int k = 0; k < len1; k++)
                {
                    if(vv[j][temp[k] - 'a'])
                    {
                        flag = 1;
                        break;
                    }
                }
                if(flag == 0)
                    sum = max(sum, a[j] * a[i]);
            }

        }

        return sum;
    }
};