letcode算法--20.螺旋矩阵

 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。

示例 1：

输入：matrix = [[1,2,3],[4,5,6],[7,8,9]]
输出：[1,2,3,6,9,8,7,4,5]

示例 2：

输入：matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12]]
输出：[1,2,3,4,8,12,11,10,9,5,6,7]

方法一：循环边界控制

class Solution {
    public List<Integer> spiralOrder(int[][] matrix) {
         int row = matrix.length;//行
         int col = matrix[0].length;//列
         int left = 0, right = col - 1,top = 0, down = row - 1;
        List<Integer> result = new ArrayList<>();
        int i = 0, j = 0;
        while (true) {
            for (j = left; j <= right; j++) {
                result.add(matrix[top][j]);
            }
            top++;
            if(top > down)break;
            for (i = top; i <= down; i++) {
                result.add(matrix[i][right]);
            }
            right--;
            if(right < left)break;
            for (j = right; j >= left; j--) {
                result.add(matrix[down][j]);
            }
            down--;
            if(down < top)break;
            for(i = down; i >= top; i --){
                result.add(matrix[i][left]);
            }
            left ++;
            if(left > right)break;
        }
        return result;
    }
}

　方法二：辅助矩阵标记

 官方题解

class Solution {
    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;
    }
}