直接枚举,时间复杂度是O(N^2L), 因为判断两个字符串是否有相同的字符,需要O(L)的时间
用位运算将字符串映射到一个int,如果完全没有重复的字符,那么这两个int &后为0
class Solution {
public:
int map(const string& s) {
int res = 0;
for (auto c : s)
res |= 1 << (c - 'a');
return res;
}
int maxProduct(vector<string>& words) {
int n = words.size();
vector<int> bitmap(n, 0);
for (int i = 0; i < n; i++)
bitmap[i] = map(words[i]);
int res = 0;
for (int i = 0; i < n; i++) {
for (int j = 0; j < i; j++) {
if ((bitmap[i] & bitmap[j]) == 0 && words[i].size() * words[j].size() > res)
res = words[i].size() * words[j].size();
}
}
return res;
}
};