Maximum Product of Word Lengths 问题及解法

问题描述：

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.

示例:

Given ["abcw", "baz", "foo", "bar", "xtfn", "abcdef"]
Return 16
The two words can be "abcw", "xtfn".

Given ["a", "ab", "abc", "d", "cd", "bcd", "abcd"]
Return 4
The two words can be "ab", "cd".

Given ["a", "aa", "aaa", "aaaa"]
Return 0
No such pair of words.

问题分析：

将每个字串中的字母都映射到二进制位上，所有出现相同字母的字串，他们对应的二进制数按位与所得的值均不为0，利用这个性质即可求解问题。

过程详见代码：

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