leetcode 318. Maximum Product of Word Lengths

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318. Maximum Product of Word Lengths

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Total Accepted: 19855 Total Submissions: 50022 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.

Credits:
Special thanks to @dietpepsi for adding this problem and creating all test cases.

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 Bit Manipulation

 

 

题解：

 

 find the maximum value of length(word[i]) * length(word[j]) where the two words do not share common letters.

重点在于 判断 有没有重复的字母

由于只有小写字母（26个），所以用 状压，位运算 （与）即可

 

 

class Solution {
public:
    int maxProduct(vector<string>& words) {
        int n = words.size();
        vector <int> len;
        vector <int> contain;
        int i,j;
        int l;
        for(i = 0;i < n;i++){
            l = words[i].length();
            len.push_back(l);
            int tmp = 0;
            for(j = 0;j < l;j++){
                int x = words[i][j] - 'a';
                tmp |= (1 << x);
            }
            contain.push_back(tmp);
        }
        int re = 0;
        for(i = 0;i < n;i++){
            for(j = i + 1;j < n;j++){
                if(contain[i] & contain[j]){
                    continue;
                }
                re = max(re,len[i] * len[j]);
            }
        }
        return re;
    }
};