[프로그래머스][C++] 조이스틱

https://programmers.co.kr/learn/courses/30/lessons/42860

코딩테스트 연습 - 조이스틱

 

글쓴이의 답

개인적인 풀이 임으로

이것보다 더 좋은 알고리즘은 많음...

이렇게도 풀이하는구나.. 공유하기 위해 올림...

#include <string>
#include <vector>
#include <iostream>

using namespace std;


int dfs(string result, string &name, int location, int count, int &answer, int moveCount) {
	if (result[location] != name[location]) {
        
		if (name[location] <= 'M') {
			while (name[location] != result[location]) {
				count++;	
                if (count >= answer) {
                    return 0;
                }
				result[location] += 1;				
                
			}
		}
		else {
			while (name[location] != result[location]) {
				count++;
                if (count >= answer) {
                    return 0;
                }
				if ('A' == result[location]) {
					result[location] = 'Z';
				}
				else {
					result[location] -= 1;
				}
			}
		}
	}

	if (result == name) {
		if (answer > count) {
			answer = count;
		}
		return 0;
	}


	count++;
	moveCount++;
    if (count >= answer) {
        return 0;
    }

	if (moveCount == result.size()) {
		return 0;
	}
	int nextLocation;
	nextLocation = location + 1;
	if (nextLocation == result.size()) {
		nextLocation = 0;
	}
	dfs(result, name, nextLocation, count, answer, moveCount);
	nextLocation = location - 1;
	if (nextLocation == -1) {
		nextLocation = result.size() - 1;
	}
	dfs(result, name, nextLocation, count, answer, moveCount);
	count--;
	moveCount--;

	return 0;

}

int solution(string name) {
	int answer = 9999999;

	string result = "";
	for (int i = 0; i < name.size(); i++) {
		result += "A";
	}
	int count = 0;
	int moveCount = 0;
	dfs(result, name, 0, count, answer, moveCount);
	dfs(result, name, 0, count, answer, moveCount);

	return answer;
}

 

 

꾸준히 하다보면 실력이 늘겠지..