Leetcode: 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.

关键是怎么计算有没有重复字符 - 位运算。想到这个之后就比较简单了，O(N^2)的时间复杂度。

class Solution {public:    int maxProduct(vector<string>& words) {        vector<int> wordsBit(words.size());        vector<int> wordsLength(words.size());        for (int i = 0; i < words.size(); ++i) {            int bits = 0;            for (int j = 0; j < words[i].size(); ++j) {                bits |= (1 << (words[i][j] - 'a'));            }            wordsBit[i] = bits;            wordsLength[i] = words[i].size();        }                int maxLength = 0;        for (int i = 0; i < wordsBit.size(); ++i) {            for (int j = i + 1; j < wordsBit.size(); ++j) {                if ((wordsBit[i] & wordsBit[j]) == 0 && wordsLength[i] * wordsLength[j] > maxLength) {                    maxLength = wordsLength[i] * wordsLength[j];                }            }        }                return maxLength;    }};