LeetCode 318: Maximum Product of Word Lengths

318. Maximum Product of Word Lengths

Difficulty: Medium
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

要找到没有相同字母的两个单词，是单词长度之积最大，因此，必须记录每个单词所含有的字母。
英文字母只有26个，可以用位来表示字母是否存在于单词中，只需要26位，直接用int类型变量储存该信息，也可以用bitset。字母存在时就将对应位的值置为1。
遍历words的每个单词，记录该单词存在哪些字母；再把记录即int型数两两相与，结果为0就表明这两个单词无相同字母；计算满足条件的单词对的长度之积，比较得出的最大值为我们要的最终结果。

代码

[c++]

class Solution {public:    int maxProduct(vector<string>& words) {        vector<int> keys(words.size(), 0);        for (int i = 0; i < words.size(); ++i) {            for (char c : words[i])                keys[i] |= 1 << (c - 'a');        }        int maxp = 0;        for (int i = 0; i < words.size(); ++i) {            for (int j = i + 1; j < words.size(); ++j) {                if ((keys[i] & keys[j]) == 0) {                    int mul = words[i].size() * words[j].size();                    maxp = maxp > mul ? maxp : mul;                }            }        }        return maxp;    }};