螺旋矩阵(中等题)

2020-11-15

54. 螺旋矩阵

难度中等769

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：

1 2	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：

1 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]

提示：

m == matrix.length
n == matrix[i].length
1 <= m, n <= 10
-100 <= matrix[i][j] <= 100

模拟

class Solution {
    public List<Integer> spiralOrder(int[][] matrix) {
        // 如果矩阵为空
        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;
        // 右，下，上，左
        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];
            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;
    }
}

作者：LeetCode-Solution
链接：https://leetcode-cn.com/problems/spiral-matrix/solution/luo-xuan-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。

按层模拟

class Solution {
    public List<Integer> spiralOrder(int[][] matrix) {
        // 如果矩阵为空
        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;
    }
}

作者：LeetCode-Solution
链接：https://leetcode-cn.com/problems/spiral-matrix/solution/luo-xuan-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。

暴力解法

class Solution {
    public List<Integer> spiralOrder(int[][] matrix) {
        int maxi = matrix.length;
        int maxj = matrix[0].length;
        int mini = 0, minj = 0;
        int indexi = -1, indexj = -1;
        List<Integer> list = new ArrayList<>();
        int maxLength = matrix.length * matrix[0].length;
        int count=0;//计数，如果没有计数，可能有的情况会多遍历几个
        while (count < maxLength) {

            for (indexi++, indexj++; indexj < maxj; indexj++) {//j增
                if(count<maxLength){
                    list.add(matrix[indexi][indexj]);
                    count++;
                }
            }
            for (indexj--, indexi++; indexi < maxi; indexi++) {//i增
                if(count<maxLength){
                    list.add(matrix[indexi][indexj]);
                    count++;
                }
            }
            maxi--;//可遍历范围-1
            maxj--;//可遍历范围-1
            mini++;//可遍历范围-1
            for (indexi--, indexj--; indexj >= minj; indexj--) {//j减
                if(count<maxLength){
                    list.add(matrix[indexi][indexj]);
                    count++;
                }
            }
            for (indexj++, indexi--; indexi >= mini; indexi--) {//i减
                if(count<maxLength){
                    list.add(matrix[indexi][indexj]);
                    count++;
                }
            }
            minj++;//可遍历范围-1
        }

        return list;
    }
}