[leetcode]318. Maximum Product of Word Lengths

最新推荐文章于 2025-04-19 00:25:33 发布

xiaocong1990

最新推荐文章于 2025-04-19 00:25:33 发布

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本文介绍了解决LeetCode上一道关于寻找两不含有相同字母的字符串并求其长度乘积的最大值的问题的方法。利用位操作技巧，通过将每个字符串映射为一个整数来快速判断两个字符串是否有共同字母。

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题目链接：https://leetcode.com/problems/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.

解题思路：字母a~z共有26个，int有32位，因此可以用int表示一个单词，如，ac的int值是0000 0000 0000 0000 0000 0000 0101，对两个数相与，得到0，说明没有相同的字母。

class Solution{
public:
    int maxProduct(vector<string>& words)
    {
        vector<int> flag;
        for(int i=0;i<words.size();i++)
        {
            int t=0;
            for(int j=0;j<words[i].size();j++)
            {
                t|=(1<<(words[i][j]-'a'));
            }
            flag.push_back(t);
        }

        int ans=0;
        for(int i=0;i<flag.size();i++)
        {
            for(int j=i+1;j<flag.size();j++)
            {
                if((flag[i]&flag[j])==0)
                    ans=max(ans,(int)(words[i].size()*words[j].size()));
            }
        }
        return ans;
    }
};