给定S串和T串,求S串中删掉任意字符后,有多少中方式能够构成T串。
开始暴力搜索超时。其实技巧没掌握好。
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struct PointS {
int index;
int count;
};
class Solution {
public:
int ans;
vector<PointS> seq;
int numDistinct(string S, string T) {
ans = 0;
if(T.length() <= 0)return ans;
seq.clear();
for(int i = 0;i < S.length();i++) {
if(S[i] == T[T.length()-1]) {
PointS item;
item.index = i;
item.count = 1;
seq.push_back(item);
}
}
matchStep(T.length() - 2,S,T,seq);
return ans;
}
private:
void matchStep (int now,string S,string T,vector<PointS>& seq) {
if(now == -1) {
for(int i = 0;i < seq.size();i++) {
PointS item = seq[i];
ans += item.count;
}
return ;
}
vector<PointS> seq_c;
seq_c.clear();
int i = 0;
int bottom = 0;
int sum = 0;
for(int i = 0;i < seq.size();i++)
sum += seq[i].count;
while(i < seq.size()) {
for(int j = bottom;j < seq[i].index;j++) {
if(S[j] == T[now]) {
PointS item;
item.index = j;
item.count = sum;
seq_c.push_back(item);
}
}
bottom = seq[i].index;
sum -= seq[i].count;
i++;
}
seq.clear();
matchStep (now - 1,S,T,seq_c);
}
};
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