Leetcode: 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].

难度：87，这道题跟 Rotate Image 很相似，都是需要把矩阵分割成层来处理，每一层都是按：1. 正上方；2. 正右方；3. 正下方；4. 正左方这种顺序添加元素到结果集合。实现中要注意两个细节，一个是因为题目中没有说明矩阵是不是方阵，因此要先判断一下行数和列数来确定螺旋的层数，mina(行， 列) /2才是螺旋数。二是有可能存在这种情况比如：3*4矩阵，螺旋只有一层，但中间还有一行元素没有没加入。所以最后要将剩余的走完。

Best solution:

public class Solution {
    public List<Integer> spiralOrder(int[][] matrix) {
        
        List<Integer> res = new ArrayList<Integer>();
        
        if (matrix.length == 0) {
            return res;
        }
        
        int rowBegin = 0;
        int rowEnd = matrix.length-1;
        int colBegin = 0;
        int colEnd = matrix[0].length - 1;
        
        while (rowBegin <= rowEnd && colBegin <= colEnd) {
            // Traverse Right
            for (int j = colBegin; j <= colEnd; j ++) {
                res.add(matrix[rowBegin][j]);
            }
            rowBegin++;
            
            // Traverse Down
            for (int j = rowBegin; j <= rowEnd; j ++) {
                res.add(matrix[j][colEnd]);
            }
            colEnd--;
            
            if (rowBegin <= rowEnd) {
                // Traverse Left
                for (int j = colEnd; j >= colBegin; j --) {
                    res.add(matrix[rowEnd][j]);
                }
            }
            rowEnd--;
            
            if (colBegin <= colEnd) {
                // Traver Up
                for (int j = rowEnd; j >= rowBegin; j --) {
                    res.add(matrix[j][colBegin]);
                }
            }
            colBegin ++;
        }
        
        return res;
    }
}

My previous solution, it's a bit tricky to handle

public class Solution {
    public List<Integer> spiralOrder(int[][] matrix) {
        ArrayList<Integer> res = new ArrayList<Integer>();
        if (matrix==null || matrix.length==0 || matrix[0].length==0) return res;
        int Min = Math.min(matrix.length, matrix[0].length);
        int levelMax = Min / 2;
        for (int level=0; level<levelMax; level++) {
            for (int i=level; i<matrix[0].length-1-level; i++) {
                res.add(matrix[level][i]);
            }
            for (int i=level; i<matrix.length-1-level; i++) {
                res.add(matrix[i][matrix[0].length-1-level]);
            }
            for (int i=matrix[0].length-1-level; i>level; i--) {
                res.add(matrix[matrix.length-1-level][i]);
            }
            for (int i=matrix.length-1-level; i>level; i--) {
                res.add(matrix[i][level]);
            }
        }
        if (Min % 2 == 1) {
            if (matrix.length < matrix[0].length) {
                for (int i=levelMax; i<matrix[0].length-levelMax; i++) {
                    res.add(matrix[levelMax][i]);
                }
            }
            else {
                for (int i=levelMax; i<matrix.length-levelMax; i++) {
                    res.add(matrix[i][levelMax]);
                }
            }
        }
        return res;
    }
}