"Leetcode" Maximum Product of Word Lengths

Given A string array words , find the maximum value of the where the words does not length(word[i]) * length(word[j]) share common letters. You may assume this each word would contain only lower case letters. If no such and words exist, return 0.

Example 1:

Given["abcw", "baz", "foo", "bar", "xtfn", "abcdef"]
Return16
The both words can be "abcw", "xtfn" .

Example 2:

Given["a", "ab", "abc", "d", "cd", "bcd", "abcd"]
Return4
The both words can be "ab", "cd" .

Example 3:

Given["a", "aa", "aaa", "aaaa"]
Return0
No such pair of words.

If this problem is compared with the violent comparison method, that is, the two-layer cycle, then the execution will time out.

You can save a key value for each word by using a Bitset method like C + +, and then compare it directly with the key value to reduce the time to compare characters when comparing words.

Like what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。

This key value is then manipulated with the key value of the other element, and the result is 0, which indicates that the word has no identical characters.

Here's the code:

classsolution (Object): Index= []    defTranschar (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']        returnL.index (c) + 1defparsewords (self,w): t=0 L= []         forIinchRange (27): L.append (0) forIinchSet (W):#Note: Using set to filter out the same characters, I initially use w directly, resulting in a run timeoutt =Self.transchar (i) l[t]= 1T=0 forIinchRange (len (l)):ifL[i] = = 1: T= t + POW (2, i)#Print W,t        returnTdefmaxproduct (self, words):""": Type WORDS:LIST[STR]: Rtype:int"""Max=0ifLen (words) = =0:return0 L= []         forIinchWords:l.append (Self.parsewords (i)) forIinchRange (len (l)): forJinchRange (i+1, Len (l)):ifL[i] & l[j] = =0:ifMax < Len (Words[i]) *Len (Words[j]): Max= Len (words[i]) *Len (words[j])returnMax

"Leetcode" Maximum Product of Word Lengths