给定一个字符串数组 words，找到 length(word[i]) * length(word[j]) 的最大值，并且这两个单词不含有公共字母。你可以认为每个单词只包含小写字母。如果不存在这样的两个单词，返回 0。
示例 1:
输入:["abcw","baz","foo","bar","xtfn","abcdef"]输出:16 解释: 这两个单词为"abcw", "xtfn"。
示例 2:
输入:["a","ab","abc","d","cd","bcd","abcd"]输出:4 解释:这两个单词为"ab", "cd"。
示例 3:
输入:["a","aa","aaa","aaaa"]输出:0 解释: 不存在这样的两个单词。
C
int maxProduct(char** words, int wordsSize)
{
int n=wordsSize;
int res=0;
int* tmp=(int*)malloc(sizeof(int)*n);
memset(tmp,0,sizeof(int)*n);
for(int i=0;i<n;i++)
{
for(int j=0;j<strlen(words[i]);j++)
{
tmp[i] |= 1<<(words[i][j]-'a');
}
}
for(int i=0;i<n;i++)
{
for(int j=i+1;j<n;j++)
{
if(0==(tmp[i]&tmp[j]))
{
int cc=strlen(words[i])*strlen(words[j]);
res=res>cc?res:cc;
}
}
}
return res;
}C++
class Solution {
public:
int maxProduct(vector<string>& words)
{
int n=words.size();
int res=0;
vector<int> tmp(n,0);
for(int i=0;i<n;i++)
{
for(int j=0;j<words[i].size();j++)
{
tmp[i] |= 1<<(words[i][j]-'a');
}
}
for(int i=0;i<n;i++)
{
for(int j=i+1;j<n;j++)
{
if(0==(tmp[i]&tmp[j]))
{
int cc=words[i].size()*words[j].size();
res=max(res,cc);
}
}
}
return res;
}
};python
class Solution:
def maxProduct(self, words):
"""
:type words: List[str]
:rtype: int
"""
n=len(words)
res=0
tmp=[0 for i in range(n)]
for i in range(n):
for j in range(len(words[i])):
tmp[i] |= 2**(ord(words[i][j])-ord('a'))
for i in range(n):
for j in range(i+1,n):
if 0==(tmp[i]&tmp[j]):
cc=len(words[i])*len(words[j])
res=max(res,cc)
return res