LeetCode 318. 最大单词长度乘积

给定一个字符串数组?words，找到?length(word[i]) * length(word[j])?的最大值，并且这两个单词不含有公共字母。你可以认为每个单词只包含小写字母。如果不存在这样的两个单词，返回 0。

示例?1:
输入: ["abcw","baz","foo","bar","xtfn","abcdef"]
输出: 16
解释: 这两个单词为 "abcw", "xtfn"。

示例 2:
输入: ["a","ab","abc","d","cd","bcd","abcd"]
输出: 4
解释: 这两个单词为 "ab", "cd"。

示例 3:
输入: ["a","aa","aaa","aaaa"]
输出: 0
解释: 不存在这样的两个单词。

class Solution:
    def maxProduct(self, words: List[str]) -> int:
        word_dict = {}
        for word in words:
            word_dict[word] = 0
            for c in word:
                word_dict[word] |= 2**(ord(c)-97)    ## 或运算、异或运算、与运算、按位取反运算
        
        ans = 0
        for i in range(len(words)-1):
            for j in range(i+1,len(words)):                
                a = words[i]
                b = words[j]
                bit_a = word_dict[a]
                bit_b = word_dict[b]
                #print(i,j,a,b,bit_a,bit_b,bit_a&bit_b)
                if bit_a&bit_b == 0:
                    ans = max(ans,len(a)*len(b))
        return ans

LeetCode 318. 最大单词长度乘积

原文：https://www.cnblogs.com/sandy-t/p/13288226.html