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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