Leetcode(54) Spiral Matrix

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Description

Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.

Example 1:

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Input:
[
 [ 1, 2, 3 ],
 [ 4, 5, 6 ],
 [ 7, 8, 9 ]
]
Output: [1,2,3,6,9,8,7,4,5]

Example 2:

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Input:
[
  [1, 2, 3, 4],
  [5, 6, 7, 8],
  [9,10,11,12]
]
Output: [1,2,3,4,8,12,11,10,9,5,6,7]

解法

这题就按照四个方向来进行数组的遍历，注意方向的转换与边界的处理，同时注意处理一行，一列为空的特殊情况。

具体代码如下：

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class Solution {
    public List<Integer> spiralOrder(int[][] matrix) {
        List<Integer> res = new ArrayList<>();
        if (matrix.length == 0) {
            return res;
        }
        if (matrix[0].length == 1) {
            for (int i = 0; i < matrix.length; i++) {
                res.add(matrix[i][0]);
            }
            return res;
        }
        if (matrix.length == 1) {
            for (int i = 0; i < matrix[0].length; i++) {
                res.add(matrix[0][i]);
            }
            return res;
        }
        int size = matrix[0].length * matrix.length;
        int i = 0;
        int j = 0;
        int upbond = 1;
        int downbond = matrix.length - 1;
        int leftbond = 0;
        int rightbond = matrix[0].length - 1;
        int dir = 0;
        res.add(matrix[0][0]);
        for (int k = 0; k < size - 1; k++) {
            switch (dir) {
                case 0: {
                    j++;
                    res.add(matrix[i][j]);
                    if (j == rightbond) {
                        dir = 1;
                        rightbond--;
                    }
                    break;
                }
                case 1: {
                    i++;
                    res.add(matrix[i][j]);
                    if (i == downbond) {
                        dir = 2;
                        downbond--;
                    }
                    break;
                }
                case 2: {
                    j--;
                    res.add(matrix[i][j]);
                    if (j == leftbond) {
                        dir = 3;
                        leftbond++;
                    }
                    break;
                }
                case 3: {
                    i--;
                    res.add(matrix[i][j]);
                    if (i == upbond) {
                        dir = 0;
                        upbond++;
                    }
                    break;
                }
            }
        }
        return res;
    }
}