318. Maximum Product of Word Lengths

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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.

思路是利用小写字母是有限的，将string变成一个int，这样更好比较。比如“abcdefghijklmnopqrstuvwxyz” => 1...1, 一共26个1。 比较的时候如果两个string有相同的letter，那么

int1&int2 !=0。然后根据这个条件向前判断，求最大值就好了。

public int MaxProduct(string[] words) {
        int res =0;
         var s =new int[words.Count()];
         for(int i=0;i< words.Count();i++)
         {
             string word = words[i];
             for(int j=0;j<word.Length;j++)
             {
                 
                 s[i] |= 1<<(word[j]-‘a‘); 
             }
             
             for(int m= i-1;m>=0;m--)
             {
                 if((s[m] & s[i]) ==0) res = Math.Max(res,words[i].Length*words[m].Length);
             }
         }
         return res;
    }

318. Maximum Product of Word Lengths

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原文地址：http://www.cnblogs.com/renyualbert/p/5902811.html