Given a binary string s, return the number of non-empty substrings that have the same number of 0’s and 1’s, and all the 0‘s and all the 1‘s in these substrings are grouped consecutively.
Substrings that occur multiple times are counted the number of times they occur.
Solution
由于连续的 /(0/) 和连续的 /(1/)， 所以我们只需要统计相邻区间 /(0,1/) 的个数，取最小即可，因为方案数就是最小个数的 /(0/1/)
点击查看代码
class Solution {
private:
int ans=0;
vector<int> cnt;
public:
int countBinarySubstrings(string s) {
int n =s.size();
int cur=1;
for(int i=0;i<n-1;i++){
if(s[i]==s[i+1])cur++;
else{
cnt.push_back(cur);cur=1;
}
}
cnt.push_back(cur);
for(int i=0;i<cnt.size()-1;i++){
//cout<<cnt[i]<<" "<<cnt[i+1]<<endl;
ans += min(cnt[i], cnt[i+1]);
}
return ans;
}
};
原创文章，作者：ItWorker，如若转载，请注明出处：https://blog.ytso.com/281956.html