剑指 Offer 29. 顺时针打印矩阵（54. 螺旋矩阵）

题目：

思路：

【1】使用辅助变量，和构建使用的偏移数据组合，然后在单次循环中对数据进行塞入

【2】不使用辅助空间，按照模拟的思维，使用四个变量记录范围值，然后按照逻辑遍历塞入数组，其中需要特殊处理的便是最后形成一行或一列的情况。

代码展示：

使用辅助变量进行单循环：

//时间3 ms击败20.1%
//内存43 MB击败88.20%
class Solution {
    public int[] spiralOrder(int[][] matrix) {
        if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
            return new int[0];
        }
        int rows = matrix.length, columns = matrix[0].length;
        boolean[][] visited = new boolean[rows][columns];
        int total = rows * columns;
        int[] order = new int[total];
        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[i] = 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;
    }
}

 

不使用辅助空间，按模拟顺序进行遍历：

//时间0 ms击败100%
//内存39.6 MB击败47.71%
//时间复杂度：O(mn)，其中 m 和 n 分别是输入矩阵的行数和列数。矩阵中的每个元素都要被访问一次。
//空间复杂度：O(1)。除了输出数组以外，空间复杂度是常数。
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;
        int left = 0, right = columns - 1, top = 0, bottom = rows - 1;
        while (left <= right && top <= bottom) {
            for (int column = left; column <= right; column++) {
                order.add(matrix[top][column]);
            }
            for (int row = top + 1; row <= bottom; row++) {
                order.add(matrix[row][right]);
            }
            if (left < right && top < bottom) {
                for (int column = right - 1; column > left; column--) {
                    order.add(matrix[bottom][column]);
                }
                for (int row = bottom; row > top; row--) {
                    order.add(matrix[row][left]);
                }
            }
            left++;
            right--;
            top++;
            bottom--;
        }
        return order;
    }
}

//时间1 ms击败61.99%
//内存43.4 MB击败44.85%
class Solution {
    public int[] spiralOrder(int[][] matrix) {
        if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
            return new int[]{};
        }
        int index=0;
        int rows = matrix.length, columns = matrix[0].length;
        int[] order = new int[rows*columns];
        int left = 0, right = columns - 1, top = 0, bottom = rows - 1;
        while (left <= right && top <= bottom) {
            for (int column = left; column <= right; column++) {
                order[index++] = matrix[top][column];
            }
            for (int row = top + 1; row <= bottom; row++) {
                order[index++] = matrix[row][right];
            }
            if (left < right && top < bottom) {
                for (int column = right - 1; column > left; column--) {
                    order[index++] = matrix[bottom][column];
                }
                for (int row = bottom; row > top; row--) {
                    order[index++] = matrix[row][left];
                }
            }
            left++;
            right--;
            top++;
            bottom--;
        }
        return order;
    }
}