容易超时,需要优化,减少重复处理次数。
class Solution {
public:
int process(string& s, vector<int>& a_indexs, char c1, char c2){
int res = 0;
for (int i=0;i<a_indexs.size();i++){
int a_index = a_indexs[i];
// find all b_index>a_index and all c_index>a_index, gain max num
int b_index = -1;
for (int j=a_index+1;j<s.size();j++){
if (s[j]==c1){
b_index = j;
break;
}
}
int c_index = -1;
for (int j=a_index+1;j<s.size();j++){
if (s[j]==c2){
c_index = j;
break;
}
}
if (b_index!=-1 && c_index!=-1){
int max = b_index>c_index?b_index:c_index;
res += s.size()-max;
}
}
return res;
}
int numberOfSubstrings(string s) {
vector<int> a_indexs;
vector<int> b_indexs;
vector<int> c_indexs;
for (int i=0;i<s.size();i++){
if (s[i]=='a'){a_indexs.push_back(i);}
if (s[i]=='b'){b_indexs.push_back(i);}
if (s[i]=='c'){c_indexs.push_back(i);}
}
//gain res
int res1 = process(s, a_indexs, 'b', 'c');
int res2 = process(s, b_indexs, 'a', 'c');
int res3 = process(s, c_indexs, 'a', 'b');
return res1+res2+res3;
}
};