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# LeetCode : 54. Spiral Matrix

2021-02-16
goodGid

## 54. Spiral Matrix

### Problem

``````Given an m x n matrix, return all elements of the matrix in spiral order.
``````

### Example

``````Input: matrix = [[1,2,3],[4,5,6],[7,8,9]]
Output: [1,2,3,6,9,8,7,4,5]
``````

### [1] Code (21. 02. 16)

``````class Solution {

int[] dx = { 0, 1, 0, -1 };
int[] dy = { 1, 0, -1, 0 };

public List<Integer> spiralOrder(int[][] matrix) {

boolean[][] isVisit = new boolean[matrix.length][matrix[0].length];

return solve(answer, matrix, isVisit, 0, 0, 0);
}

private List<Integer> solve(List<Integer> answer, int[][] matrix, boolean[][] isVisit,
int dir, int x, int y) {

isVisit[x][y] = true;

int nx = x + dx[dir];
int ny = y + dy[dir];

int nextDir = getNextDirection(dir + 1);
int nx2 = x + dx[nextDir];
int ny2 = y + dy[nextDir];

if (isInRange(matrix, nx, ny)
&& isVisit[nx][ny] == false) {
solve(answer, matrix, isVisit, dir, nx, ny);

} else if (isInRange(matrix, nx2, ny2)
&& isVisit[nx2][ny2] == false) {
solve(answer, matrix, isVisit, nextDir, nx2, ny2);
}

}

private int getNextDirection(int dir) {
return dir % 4;
}

private boolean isInRange(int[][] matrix, int x, int y) {
return x >= 0 && x < matrix.length && y >= 0 && y < matrix[0].length;
}
}
``````

Check Point

1. 달팽이 모양으로 순회하면서 인자 값 출력

2. input 값에 대해서는 고려하지 않아도 됨

3. 코너 케이스는 따로 없어 보임

Algorithm Description

• 단순 구현문제