Leetcode 318. Maximum Product of Word Lengths

-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-
第一次首先暴力搜索一遍啦啦啦，大循环需要比较每两个字符串是否有相同字符同时计算每两个字符串之间的长度乘积，需要复杂度为O(nlgn)的循环，于是算法的瓶颈在于如何快速比较出两个字符串是否有相同字符。
暴力的方法是对于字符串a,b，做一个length(a)*length(b)的循环对每一个a的字符在b中找是否有匹配，找到相同字符则判断出两个字符串含有相同字符。
试了一遍，暴力果然不能过。于是想用一种方式来表示每个字符串中字母的存在。因此在这里用一个数的二进制位来标识对应字母是否存在。比如第一位为1表示a存在于这个字符串中。然后用两个字符串的数字表示的与来判断是否含有相同的字符。
-代码-

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