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  • 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;
        }
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  • 原文地址:https://www.cnblogs.com/renyualbert/p/5902811.html
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