给定一个包含 m x n 个元素的矩阵（m 行, n 列），请按照顺时针螺旋顺序，返回矩阵中的所有元素。

示例 1:

输入:
[
[ 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:
[
[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]

class Solution {
  public List<Integer> spiralOrder(int[][] matrix) {
    List<Integer> result = new LinkedList<>();
    if(matrix.length==0)  return result;
    int upBound = 0;
    int rightBound = matrix[0].length-1;
    int leftBound = 0;
    int downBound = matrix.length-1;
    while(true){
      for(int i=leftBound; i<=rightBound; ++i)  
        result.add(matrix[upBound][i]);
      if(++upBound>downBound) break;
      for(int i=upBound; i<=downBound; ++i)  
        result.add(matrix[i][rightBound]);
      if(--rightBound<leftBound) break;
      for(int i=rightBound; i>=leftBound; --i)  
        result.add(matrix[downBound][i]);
      if(--downBound<upBound) break;
      for(int i=downBound; i>=upBound; --i)  
        result.add(matrix[i][leftBound]);
      if(++leftBound>rightBound) break;
    }
    return result;
  }
}

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