leetcode刷题记录20

发表于   |   分类于

介绍

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.

思路

位操作 利用集合去重

代码

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
class Solution(object):
    def maxProduct(self, words):
        """
        :type words: List[str]
        :rtype: int
        """
        wordlength=[len(i) for i in words]
        l=len(words)
        if l==0:
            return 0
        else:
            r=0
            wordlist=[list(set(i)) for i in words]
            wordnumbers=[0 for i in range(l)]
            for i in range(l):
                numsum=0
                for j in wordlist[i]:
                     numsum=numsum+2**(ord(j)-97)
                wordnumbers[i]=numsum
            for i in range(l):
                if i==l-1:
                    return r
                else:
                    for j in range(i+1,l):
                        if wordnumbers[i]^wordnumbers[j]==wordnumbers[i]+wordnumbers[j]:
                             r=max(wordlength[i]*wordlength[j],r)