318. Maximum Product of Word Lengths

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CarryPotMan

最新推荐文章于 2024-12-13 13:09:19 发布

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分类专栏：算法 leetcode

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本文介绍了一种使用位操作来解决LeetCode上一道关于寻找两个不含有相同字母的最大长度字符串乘积的问题的方法。通过巧妙利用位运算，可以有效地检查两字符串是否有共同字符，并最终找到满足条件的最大乘积。

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.

https://leetcode.com/discuss/74580/bit-shorter-c

判断两个字符串有没有共同的字符（都是小写字母）

class Solution {
public: 
    int maxProduct(vector<string>& words) {
        int result = 0;
        if(words.empty()) return result;
        unordered_map<int ,int> maps;
        for(auto word : words){
            int mask = 0;
            for(auto c : word)
                mask |= 1<<(c-'a');   //这句是精华，用比特位指示出现某个字符
            maps[mask] = max(maps[mask],(int) word.size());         
        }
        for(auto a : maps)
            for(auto b : maps){
                if(!(a.first & b.first))
                    result = max(result, a.second*b.second);
            }
        return result;
    }
};