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;
}
}
