题解 | #吐泡泡#

思路

法1：手写栈

Code

#include <bits/stdc++.h>

using namespace std;

const int N = 110;

char stk[N];
string s;
int hh;

bool check(int h,int i){
    if(stk[i]=='o'&&stk[h]=='o') return true;
    if(stk[i]=='O'&&stk[h]=='O') return true;
    return false;
}

int main(){
    while(cin>> s){
        hh=0;
        for(int i=0;i<s.size();i++){
            stk[++hh]=s[i];
          while(hh>1&&check(hh,hh-1)){
            if(stk[hh]=='o') hh--,stk[hh]='O';
            else hh-=2;
          } 
        }
        for(int i=1;i<=hh;i++) cout<<stk[i];
        puts("");
        // hh不再是0,所以需要重新置零
    }
    return 0;
}

法2：stl 栈

Code

#include <bits/stdc++.h>

using namespace std;

const int N = 110;

string s;

bool check(char c1,char c2){
    if(c1==c2) return true;
    return false;
}

int main(){
    while(cin>> s){
        stack<char> stk;
        for(int i=0;i<s.size();i++){
           stk.push(s[i]);
           while(stk.size()>=2){
               char c1=stk.top(); stk.pop();
               char c2=stk.top(); stk.pop();

               stk.push(c2);  stk.push(c1);

               if(!check(c1,c2)) break;

               if(c1=='o'){
                    stk.pop();  stk.pop(); 
                    stk.push('O');
               }
               else  stk.pop(),stk.pop(); 
           } 
        }

        string res;
        while(stk.size()){
            res+=stk.top();
            stk.pop();
        }
        reverse(res.begin(),res.end()); 
        cout<<res<<endl;
       // hh不再是0,所以需要重新置零
    }

    return 0;
}