顺时针打印矩阵

2020-12-17

输入一个矩阵，按照从外向里以顺时针的顺序依次打印出每一个数字。

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]
 

限制：

0 <= matrix.length <= 100
0 <= matrix[i].length <= 100
注意：本题与主站 54 题相同：https://leetcode-cn.com/problems/spiral-matrix/

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/shun-shi-zhen-da-yin-ju-zhen-lcof
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

class Solution {
    public int[] spiralOrder(int[][] matrix) {
        if (matrix.length == 0 || matrix[0].length == 0) return new int[]{};
        int startY = 0, startX = 0;
        int endY = matrix.length, endX = matrix[0].length;

        int sum = matrix.length * matrix[0].length;
        int count = 0;

        int[] result = new int[sum];

        while (count < sum) {

            for (int i = startX; i < endX; i++) {//向右遍历
                result[count] = matrix[startY][i];
                count++;
            }
            if (count >= sum) return result;
            for (int i = startY + 1; i < endY; i++) {//向下遍历
                result[count] = matrix[i][endX - 1];
                count++;
            }
            if (count >= sum) return result;

            for (int i = endX - 1 - 1; i >= startX; i--) {//向左遍历
                result[count] = matrix[endY - 1][i];
                count++;
            }
            if (count >= sum) return result;

            for (int i = endY - 1 - 1; i > startY; i--) {//向上遍历
                result[count] = matrix[i][startX];
                count++;
            }
            startX++;
            endX--;
            startY++;
            endY--;
        }
        return result;
    }
}

精妙绝仑的模拟


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];
            int 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;
    }
}

按层模拟

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;
        int[] order = new int[rows * columns];
        int index = 0;
        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;
    }
}

作者：LeetCode-Solution
链接：https://leetcode-cn.com/problems/shun-shi-zhen-da-yin-ju-zhen-lcof/solution/shun-shi-zhen-da-yin-ju-zhen-by-leetcode-solution/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。