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 = ;i < n;i++){
             l = words[i].length();
             len.push_back(l);
             int tmp = ;
             for(j = ;j < l;j++){
                 int x = words[i][j] - 'a';
                 tmp |= ( << x);
             }
             contain.push_back(tmp);
         }
         int re = ;
         for(i = ;i < n;i++){
             for(j = i + ;j < n;j++){
                 if(contain[i] & contain[j]){
                     continue;
                 }
                 re = max(re,len[i] * len[j]);
             }
         }
         return re;
     }
 };