LeetCode: Maximum Product of Word Lengths

Problem:

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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Solutuion: 关键词 位操作

class Solution {
public:
    int maxProduct(vector<string>& words) {
       
      //  解题思路：用一个26位来表示每个字符串字母出现的情况 若出现则字母对应的为为1  前期预处理过程 
      //本题只关心字母是否出现过
   
      if(words.empty() || words.size()==0)
        return 0;
           
      //字母预处理
      vector<int> processwords(words.size(),0);
      
      for(int i=0;i<words.size();i++)
      {
          string temp_str=words[i];
          for(int j=0;j<temp_str.length();j++)
            processwords[i]|=1<<(temp_str[j]-'a');
          
      }
      
      int maxproduct=0;
      for(int i=0;i<words.size()-1;i++)
        for(int j=i+1;j<words.size();j++)
            if((processwords[i]&processwords[j])==0)
            {
              int len=words[i].length()*words[j].length();
               maxproduct=max(maxproduct,len);
            }
    
        return maxproduct;
        
        
       
        
    }
};