Leetcode54 螺旋矩阵

题目描述

给你一个 m 行 n 列的矩阵 matrix ，请按照 顺时针螺旋顺序 ，返回矩阵中的所有元素。

解法一

按照上图箭头画的顺序，一圈一圈的遍历

定义四个变量，来控制遍历过程中的边界

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

        List<Integer> res = new ArrayList<Integer>();

        if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
            return res;
        }

        int rows = matrix.length, columns = matrix[0].length;
        int left = 0, right = columns - 1, top = 0, bottom = rows - 1;
        while (left <= right && top <= bottom) {

            for (int column = left; column <= right; column++) {
                res.add(matrix[top][column]);
            }

            for (int row = top + 1; row <= bottom; row++) {
                res.add(matrix[row][right]);
            }

            if (left < right && top < bottom) { // 防止一行或一列的情况
                for (int column = right - 1; column > left; column--) {
                    res.add(matrix[bottom][column]);
                }
                for (int row = bottom; row > top; row--) {
                    res.add(matrix[row][left]);
                }
            }
            left++;
            right--;
            top++;
            bottom--;
        }
        return res;
    }

方法二：

比较推荐这种方式，比较好理解

public List<Integer> spiralOrder(int[][] matrix) {
    List<Integer> order = new ArrayList<Integer>();
    if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
        return order;
    }
    int rows = matrix.length, columns = matrix[0].length;
    boolean[][] visited = new boolean[rows][columns];
    int total = rows * columns;
    int row = 0, column = 0;
    int[][] directions = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
    int directionIndex = 0;
    for (int i = 0; i < total; i++) {
        order.add(matrix[row][column]);
        visited[row][column] = true;
        int nextRow = row + directions[directionIndex][0], nextColumn = column + directions[directionIndex][1];
        if (nextRow < 0 || nextRow >= rows || nextColumn < 0 || nextColumn >= columns || visited[nextRow][nextColumn]) {
            directionIndex = (directionIndex + 1) % 4;
        }
        row += directions[directionIndex][0];
        column += directions[directionIndex][1];
    }
    return order;
}

方法三：递归

 public static int[][] directions = {{0,1},{1,0},{0, -1},{-1, 0}};
    //public static int direction = 0; 定义为static,在多次运行后，某一次开始是direction可能不是从0开始，所以测试用例报错了
    public int direction = 0;
    //定义一个标记矩阵，每次遍历完 matrix 的方格，用标记矩阵标记
    public boolean[][] seen;

    public List<Integer> spiralOrder(int[][] matrix) {
        List<Integer> res = new ArrayList<>();
        int rows = matrix.length;
        int cols = matrix[0].length;
        seen = new boolean[rows][cols];
        dfs(res, matrix, rows, cols, 0, 0);

        return res;
    }
    public void dfs(List<Integer> res, int[][] matrix, int rowNums, int colNums, int x, int y) {
        //判断条件：超出边界，或者遇到已经赋值过的方格
        if (x < 0 || x >= rowNums || y < 0 || y >= colNums || seen[x][y] == true) {
            //撞到墙了，需要转向
            direction = (direction + 1) % 4;
            return;
        }
        //标记该方格
        seen[x][y] = true;
        //添加该方格进数组
        res.add(matrix[x][y]);
        //循环供一次原方向，一次转向
        for (int i = 0; i < 2; i++) {
            int mx = x + directions[direction][0];
            int my = y + directions[direction][1];
            //递归下一个方格
            dfs(res, matrix, rowNums, colNums, mx, my);
        }
    }

参考代码：官方题解