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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.
此题如果用暴力比较法，也就是两层循环依次比较的话，执行会超时。
可以通过类似C++中bitset的方法来保存每个单词的一个key值，然后直接用key值进行比较，减少单词之间比较的时候比较字符的时间。
比如单词abcw，我们可以创建一个l = [0,0,0.....0] 有27个0的数组，并按26个字母的顺序依次给a-z索引为1~26，最后把a,b,c,w四位对应的元素的值置为1，计算 pow(2,1)+pow(2,2)+pow(2,3)+pow(2,23)的和即为这个元素的key值。
再用这个key值与其他元素的key值做与操作，结果为0，则表示单词无相同的字符。
下面是代码：
class Solution(object): index = [] def transChar(self,c): l = [‘a‘,‘b‘,‘c‘,‘d‘,‘e‘,‘f‘,‘g‘,‘h‘,‘i‘,‘j‘,‘k‘,‘l‘,‘m‘,‘n‘,‘o‘,‘p‘,‘q‘,‘r‘,‘s‘,‘t‘,‘u‘,‘v‘,‘w‘,‘x‘,‘y‘,‘z‘] return l.index(c) + 1 def parseWords(self,w): t = 0 l = [] for i in range(27): l.append(0) for i in set(w): #注：这里用set过滤掉相同的字符，我最初直接用w，导致运行超时 t = self.transChar(i) l[t] = 1 t = 0 for i in range(len(l)): if l[i] == 1: t = t + pow(2,i) #print w,t return t def maxProduct(self, words): """ :type words: List[str] :rtype: int """ max = 0 if len(words) == 0: return 0 l = [] for i in words: l.append(self.parseWords(i)) for i in range(len(l)): for j in range(i+1,len(l)): if l[i] & l[j] == 0: if max < len(words[i]) * len(words[j]): max = len(words[i]) * len(words[j]) return max
【leetcode】Maximum Product of Word Lengths
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原文地址：http://www.cnblogs.com/seyjs/p/5133241.html
