leetcode 318: Maximum Product of Word Lengths

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.

题意：求，不包含相同字符的两个string的最大长度积；

思路：

如何降低判断两string是否包含相同字符的开销？？？

用bit来表示string元素中a-z每个字符是否出现；

class Solution {
public:
int maxProduct(vector<string>& words) {
int len = words.size();
if(len==0)
return 0;
vector<int> v(len,0);
vector<int> l(len,0);
for(int i=0;i<len;i++)
{
string tstr = words[i];
int tlen = tstr.length();
l[i] = tlen;
for(int j=0;j<tlen;j++)
{
int ttmp = 1<<(tstr[j]-'a');
v[i] |= ttmp;
}
}

int ans = 0;
for(int i=0;i<len;i++)
{
for(int j=i+1;j<len;j++)
{
int t2 = v[i]&v[j];
if(t2!=0)
continue;
int tmp = l[i]*l[j];
if(tmp>ans)
ans = tmp;
}
}
return ans;
}
};

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