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.
根据题目的描述,我们可知,最耗时的地方在于判断两个字符串中是否存在重复的字符。
一提到判断重复字符,脑海中马上浮现哈希大法。
又看到题目要保证字符串均为小写字母,也就说总共26个字母,我们需要26位,一个int足矣。
我们自定义,如果某字符出现,该bit置1,否则置0.
则比较两个字符串是否有重复字符,就是让这两个标志int 进行与操作,结果为0则表明没有重复字符。
好的,本题的难点已结束。code如下:
易知时间复杂度为O(n2),空间复杂度为O(n),其中n为所有字符串的个数。
class Solution {
public:
int maxProduct(vector<string>& words) {
int isDuplicate[2000] = {0};
int nSize = words.size();
for(int i = 0; i < nSize; i++)
{
int length = words[i].length();
for(int j = 0; j < length; j++)
isDuplicate[i] |= 1 << (words[i][j] - 'a');
}
int maxMul = 0, tempMul = 0;
for(int i = 0; i < nSize -1; i++)
for(int j = i +1; j < nSize; j++)
if((isDuplicate[i] & isDuplicate[j]) == 0)
{
tempMul = words[i].length()*words[j].length();
if(maxMul < tempMul)
maxMul = tempMul;
}
return maxMul;
}
};
Submission Details
174 / 174 test cases passed.
Status:
Accepted
Runtime: 104 ms
Submitted: 0 minutes ago