LeetCode #318 - Maximum Product of Word Lengths

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本文介绍了一种高效算法，用于解决寻找字符串数组中两单词不包含相同字母时的最大长度乘积问题。通过使用int类型的26低位来表示26个字母的出现情况，实现了快速判断两个单词是否有共同字母的目标。

题目描述：

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:

Input: ["abcw","baz","foo","bar","xtfn","abcdef"]

Output: 16 
Explanation: The two words can be "abcw", "xtfn"

Example 2:

Input: ["a","ab","abc","d","cd","bcd","abcd"]

Output: 4 
Explanation: The two words can be "ab", "cd".

Example 3:

Input: ["a","aa","aaa","aaaa"]

Output: 0 
Explanation: No such pair of words.

对于任意一对不含相同字母的单词，求它们的单词长度乘积的最大值。这道题如果用哈希表来统计每个单词中26个字母的出现情况，空间复杂度较高，而且如果需要判断任意两个单词是否含有相同字母也比较麻烦，考虑到int类型有32位，用较低的26位来存储每个单词的26个字母的出现情况正好合适，而且只需要与操作就能判断两个单词中是否含有相同字母，非常简便。

class Solution {
public:
    int maxProduct(vector<string>& words) {
        int n=words.size();
        vector<int> code(n,0);
        int result=0;
        for(int i=0;i<n;i++) code[i]=encode(words[i]);
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<n;j++)
            {
                if(i!=j&&(code[i]&code[j])==0) 
                {
                    int mul=words[i].size()*words[j].size();
                    result=max(result,mul);
                }
            }
        }
        return result;
    }
    
    int encode(string s)
    {
        int result=0;
        for(int i=0;i<s.size();i++) result|=1<<(s[i]-'a');
        return result;
    }
};