leetcode minimum window substring
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class Solution {public: bool findAll(vector<int> to_find,vector<int> has_find){bool flag=true;for(int i=0;i<256;i++){if(to_find[T[i]]<=has_find[T[i]]){flag=flag&&true;}else{flag=false;break;}}return flag;} string minWindow(string S, string T) { // Start typing your C/C++ solution below // DO NOT write int main() functionint lenT=T.length();int lenS=S.length();if(lenT==0)return "";if(lenS==0)return ""; vector<int> to_find(256);vector<int> has_find(256);for(int i=0;i<lenT;i++){to_find[T[i]]++;}int left=0;int minleft=0;int minLength=INT_MAX;for(int i=0;i<lenS;i++){if(to_find[S[i]]>0){has_find[S[i]]++;//cout<<S[i];if(findAll(to_find,has_find)){//window contains all chars in T,then move left//assert(S[i]=='b');//cout<<"in"<<endl;//cout<<S[i]<<endl;while(true){//move left//cout<<has_find[S[left]]<<endl;//cout<<to_find[S[left]]<<endl;if(has_find[S[left]]<=to_find[S[left]]&&has_find[S[left]]!=0){//condition is left is redundantbreak;}//cout<<"in"<<endl;if(has_find[S[left]]!=0)has_find[S[left]]--;//left++;}if(minLength>(i-left+1)){//see if minlengthminLength=i-left+1;minleft=left;}//has_find[S[left]]--;//left++;//cout<<S[left]<<endl;}}}if(minLength==INT_MAX){return "";}else{//cout<<minleft<<minLength;return S.substr(minleft,minLength);} }};
