프로그래머스 햄버거만들기 (아직푸는중)

링크 : https://school.programmers.co.kr/learn/courses/30/lessons/133502?language=cpp 

프로그래머스

[풀이1 스트링으로?]

#include <string>
#include <vector>
#include <bits/stdc++.h>

using namespace std;
const int bread = 1;
const int vege = 2;
const int meat = 3;

int solution(vector<int> ingredient) {
    int answer = 0;
    
    stringstream ss;
    copy(ingredient.begin(), ingredient.end(), ostream_iterator<int>(ss, ""));
    string s = ss.str();
    const string buger = "1231";
    int findidx =0;

    do
    {
         findidx = s.find("1231");
        if(findidx == -1)
        {
            break;
        }
        s.erase(findidx, 4);
        answer++;
       

    } while(findidx != -1);
    
    return answer;
}

개같이 실패~

 

 

 

 

 

 

 

split 잘 쓰면 시간복잡도 더 줄일 수 있을테넫

#include <string>
#include <vector>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <regex>
#include <queue>

using namespace std;

int solution(vector<int> ingredient) {
	int answer = 0;
	stringstream ss;
	copy(ingredient.begin(), ingredient.end(), ostream_iterator<int>(ss, ""));
	string s = ss.str();
	const string buger = "1231";
	int findidx = 0;

	queue<string> bugerQueue;
	findidx = s.find(buger);
	if (findidx == -1)
	{
		return 0;
	}

	bugerQueue.push(s.substr(findidx + 4));//erase(findidx, 4);
	string fronts = s.substr(0, findidx);
	answer++;

	do
	{
		fronts.append(bugerQueue.front());

		findidx = fronts.find(buger);
		if (findidx == -1)
		{
			fronts = bugerQueue.front();

		}
		else
		{
			int orginsize = fronts.size();

			bugerQueue.pop();
			if(findidx > 4) bugerQueue.push(fronts.substr(0, findidx));
			if(orginsize - findidx > 4) bugerQueue.push(fronts.substr(findidx + 4));//erase(findidx, 4);
			answer++;
			continue;
		}
		bugerQueue.pop();

	} while (bugerQueue.empty() == false);

	return answer;
}

int main()
{
	solution({ 2, 1, 1, 2, 3, 1, 2, 3, 1 });
	return 0;
}

오히려 erase보다 점수가 더 안좋아졌다