Spiral Matrix

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

For example, Given the following matrix:

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

You should return [1,2,3,6,9,8,7,4,5].

Tips:

回旋增加，顺序是向右，向下，向左，向上为一轮。注意判断边界条件即可。

There are 4 modes of changing index. It can go up, down, left or right. Use an array to store the boundaries of 4 modes. Read out one number each time and change the index based on mode and boundaries.

Code:

public class Solution {
    public List<Integer> spiralOrder(int[][] matrix) {
        List<Integer> result = new ArrayList<>();
        if (matrix == null || matrix.length == 0) {
            return result;
        }
        int left = 0;
        int right = matrix[0].length - 1;
        int top = 0;
        int bottom = matrix.length -1;
        while (left <= right && top <= bottom) {
            // traverse right
            for (int i = left; i <= right; i++) {
                result.add(matrix[top][i]);
            }
            top++;
            // traverse down
            for (int j = top; j <= bottom; j++) {
                result.add(matrix[j][right]);
            }
            right--;
            // traverse left: should judge first 
            // after changes it can be left > right || top > bottom,
            // should return in this case
            if (top <= bottom) {
                for (int i = right; i >= left; i--) {
                    result.add(matrix[bottom][i]);
                }
                bottom--;
            }
            // traver right: also should judge first
            if (left <= right) {
                for (int j = bottom; j >= top; j--) {
                    result.add(matrix[j][left]);
                }
                left++;
            }
        }
        return result;
    }
}