zoukankan      html  css  js  c++  java
  • 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

    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.

    Subscribe to see which companies asked this question

    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
  • 相关阅读:
    c++ 中bool 的默认值
    cocos2d CCLOG格式符号表
    c++数组指针bug
    cocos2d-x-2.2.6创建工程
    Nape实现坐标旋转角度回弹
    haxe 中使用音效
    haxe 嵌入swf 读取里面的内容
    haxe 配置
    Spring Tool Suite(STS)基本安装配置
    git提交忽略文件.gitignore内容
  • 原文地址:https://www.cnblogs.com/bonelee/p/6193580.html
Copyright © 2011-2022 走看看