重复删除子字符串“10”中的字符形成的字典序最小的字符串

#include <bits/stdc++.h>
using namespace std;
// function to get the lexicographically smallest string 
string lexSmall(string str){
   string ans = ""; // string to store the answer 
   bool flag = true; // variable to mark the changes done     
   while(flag){
      flag = false;
      int len = str.length(); // getting size of the string         
      for(int i=0; i<len; i++){
         if(str[i] == '0'){
            ans += str[i];
         }
         else if(i != len-1 && str[i+1] == '0'){
            // current index is 1 and next is 0
            // so skip this and mark the flat true 
            // indicating the change in the string 
            flag = true;
            continue;
         }
         else{
            ans += str[i];
         }
      }
      if(flag){
         str = ans;
         ans = "";
      }
   }
   return ans;
}
int main(){
   string str = "1101010011"; // given string     
   // calling the function to get the required string 
   cout<<"The smallest string after a certain number of changes is "<<lexSmall(str)<<endl;
   return 0;
}

#include <bits/stdc++.h>
using namespace std;
// function to get the lexicographically smallest string 
string lexSmall(string& str){
   string ans = ""; // string to store the answer 
   bool flag = false; // variable to mark the zero
   int len = str.length(); // getting length of the string     
   for(int i=len-1 ; i >= 0; i--){
      if(str[i] == '0'){
         ans += str[i];
         flag = true;
      }
      else if(flag){
         // flag is true means earlier there is a zero present 
         // so, no need to add this current character 
         continue;
      }
      else{
         ans += str[i];
      }
   }
   // reversing the current string 
   reverse(ans.begin(), ans.end());    
   return ans;
}
int main(){
   string str = "1101010011"; // given string     
   // calling the function to get the required string 
   cout<<"The smallest string after a certain number of changes is "<<lexSmall(str)<<endl;
   return 0;
}