[프로그래머스] 문자열, 사칙연산, 시뮬레이션, 2차원 배열

• 숨어잇는 숫자의 덧셈 (2)

#include <string>
#include <vector>

using namespace std;

int solution(string my_string) {
    int answer = 0, n=0;
    for(int i=0; i< my_string.size(); i++){
        if(isdigit(my_string[i])) n = n*10 + my_string[i]-'0';
        else{
            answer += n;
            n =0;
        }
    }
    answer += n;
    return answer;
}

• 안전지대

#include <string>
#include <vector>

using namespace std;

vector<int> x={-1,0,1,-1,1,-1,0,1};
vector<int> y={1,1,1,0,0,-1,-1,-1};

int solution(vector<vector<int>> board) {
    int answer = 0;
    for(int i=0; i< board.size(); i++){
        for(int j=0; j<board[i].size(); j++){
            if(board[i][j]==1){
                answer++;
                board[i][j]++;
                for(int k=0; k<8; k++){
                    int dx = i + x[k];
                    int dy = j + y[k];
                    
                    if((-1 < dx && dx < board.size())
                       &&(-1 < dy && dy < board.size())
                       &&(board[dx][dy]==0)){
                        board[dx][dy] = 2;
                        answer++;
                    }
                }
            }
        }
    }
    return board.size()*board.size() - answer;
}

• 삼각형의 완성 조건 (2)

#include <string>
#include <vector>

using namespace std;

int solution(vector<int> sides) {
    if(sides[0] > sides[1]) swap(sides[0],sides[1]);
    return sides[0] * 2 -1;
}

• 외계어 사전

#include <string>
#include <vector>

using namespace std;

int solution(vector<string> spell, vector<string> dic) {
    int answer = 2;
    int arr[26];
    int check[26];
    for(int i=0; i< spell.size(); i++){
        arr[spell[i][0]-'a']=1;
    }
    for(int i=0; i<dic.size(); i++){
        int num=0;
        for(int j=0; j<dic[i].size(); j++){
            if(arr[dic[i][j]-'a']==1 && check[dic[i][j]-'a']==0){
                check[dic[i][j]-'a']=1;
                num++;
            }
            else break;
        }
        if(num==spell.size()){
            answer =1;
            break;
        }
        for(int j=0; j<26; j++){
            check[j]=0;
        }
    }
    return answer;
}