318. Maximum Product of Word Lengths

最新推荐文章于 2025-09-05 23:16:08 发布

无差别刷题

最新推荐文章于 2025-09-05 23:16:08 发布

阅读量122

点赞数

CC 4.0 BY-SA版权

文章标签： bit manipulation leetcode

本文链接：https://blog.youkuaiyun.com/wilzxu/article/details/88278147

本文介绍了一种使用位操作的算法，旨在找到字符串数组中两个不包含相同字母的单词，返回它们长度的乘积。通过示例展示了如何利用位运算判断两单词是否符合要求，并给出具体实现代码。

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.

方法1: bit manipulation

思路：
将每一个字符是否出现用一个int (32 bits) 来表示。比如如果a出现，第一位是1，b出现，第二位是1。当两个单词的 & 运算结果为0 时，说明没有相同的字符出现，此时可以计算长度积并更新最大值。

易错点：

位与运算： &
! (mask[i] & mask[j]) ，要用（）来提高优先级
size () * size()要转化成int

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