LeetCode 【318. 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.

思路：对每一个元素都进行掩码设置，因为一共就26个字母，所以1个int的bit位就可以代表一个字母，比如"a"就可以表示成0000 0000 0000 0000 0000 0000 0000 0001,"bc"可以表示成0000 0000 0000 0000 0000 0000 0000 0110；如果两个元素的掩码位与以后结果为0，表示他们之间没有相同的字母，则乘积就是两串字符串的长度的乘积。

代码

class Solution {
public:
    int maxProduct(vector<string>& words) {
        vector<int> bitMask(words.size(), 0);
        for( int i = 0; i < words.size(); i++ ){
            for( int j = 0; j < words[i].size(); j++ ){
                bitMask[i] |= 1<<(words[i][j]-'a');
            }
        }
        int maxProduct = 0;
       for( int i = 0; i < words.size(); i++ ){
           for( int j = i + 1; j < words.size(); j++ ){
               if( (bitMask[i]&bitMask[j] )== 0){
               	int len1 = words[i].size();
               	int len2 = words[j].size();
                maxProduct = max(maxProduct, (len1*len2));
               }
           }
       } 
       return maxProduct;
    }
};