1. 생각
일단 k번 반복해야 한다. -> for(k)
초기위치 (0,0)
방향
0 : 오른쪽
1 : 아래
2 : 왼쪽
3 : 위
조건에 따라 이동
2. 생각대로 코딩 [ 코드 ] public int[] solution(int[][] board, int k){
int[] answer = new int[2];
//초기화
answer[0] = 0;
answer[1] = 0;
int direction = 0;
for (int i = 0; i < k; i++) {
boolean change = false; //방향전화 변수
// 넘어가면 방향 전환
if (direction == 0 && answer[1] == board.length-1) {change = true;}
if (direction == 1 && answer[0] == board.length-1) {change = true;}
if (direction == 2 && answer[1] == 0) {change = true;}
if (direction == 3 && answer[0] ==0 ) {change = true;}
// 이동
if(!change) {
if (direction == 0 && answer[1] != board.length-1) {
if(board[answer[0]][answer[1]+1] == 0) {
answer[1]++;
} else {
change = true;
}
}
if (direction == 1 && answer[0] != board.length-1) {
if(board[answer[0]+1][answer[1]] == 0) {
answer[0]++;
} else {
change = true;
}
}
if (direction == 2 && answer[1] != 0) {
if(board[answer[0]][answer[1]-1] == 0) {
answer[1]--;
} else {
change = true;
}
}
if (direction == 3 && answer[0] !=0 ) {
if(board[answer[0]-1][answer[1]] == 0) {
answer[0]--;
} else {
change = true;
}
}
}
// 방향전환
if(change) {
if(direction<=2) {
direction++;
} else {
direction=0;
}
}
}
return answer;
}
public static void main(String[] args){
Problem02 T = new Problem02();
int[][] arr1 = {{0, 0, 0, 0, 0},
{0, 1, 1, 0, 0},
{0, 0, 0, 0, 0},
{1, 0, 1, 0, 1},
{0, 0, 0, 0, 0}};
System.out.println(Arrays.toString(T.solution(arr1, 10)));
int[][] arr2 = {{0, 0, 0, 1, 0, 1},
{0, 0, 0, 0, 0, 0},
{0, 0, 0, 0, 0, 1},
{1, 1, 0, 0, 1, 0},
{0, 0, 0, 0, 0, 0},
{0, 0, 0, 0, 0, 0}};
System.out.println(Arrays.toString(T.solution(arr2, 20)));
int[][] arr3 = {{0, 0, 1, 0, 0},
{0, 1, 0, 0, 0},
{0, 0, 0, 0, 0},
{1, 0, 0, 0, 1},
{0, 0, 0, 0, 0}};
System.out.println(Arrays.toString(T.solution(arr3, 25)));
}[ 결과 ] [2, 2]
[4, 5]
[0, 1]
3. 강의 [ 코드 ] public int[] solution2(int[][] board, int k){
int[] answer = new int[2];
int n = board.length;
int[] dx = {-1, 0, 1, 0};
int[] dy = {0, 1, 0, -1};
int x = 0, y = 0, d = 1, count = 0;
while(count < k){
count++;
int nx = x + dx[d];
int ny = y + dy[d];
if(nx < 0 || nx >= n || ny < 0 || ny >= n || board[nx][ny] == 1){
d = (d + 1) % 4;
continue;
}
x = nx;
y = ny;
}
answer[0] = x;
answer[1] = y;
return answer;
}[ 차이 ]
