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318. Maximum Product of Word Lengths ——本质:英文单词中字符是否出现可以用26bit的整数表示

时间:2016-12-18 01:27:08      阅读:146      评论:0      收藏:0      [点我收藏+]

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.

Credits:
Special thanks to @dietpepsi for adding this problem and creating all test cases.

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class Solution(object):
    def maxProduct(self, words):
        bits_words = [reduce(lambda s,x: s|(1<<ord(x)-ord(a)), w, 0) for w in words]
        ans = 0
        for i, word in enumerate(words):
            for j in range(i+1, len(words)):
                if len(word)*len(words[j]) > ans and self.has_diff(bits_words[i], bits_words[j]):
                    ans = len(word)*len(words[j])
        return ans
    
    def has_diff(self, w1, w2):
        return (w1&w2) == 0

 

318. Maximum Product of Word Lengths ——本质:英文单词中字符是否出现可以用26bit的整数表示

原文:http://www.cnblogs.com/bonelee/p/6193580.html

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