顺时针打印矩阵
输入一个矩阵，按照从外向里以顺时针的顺序依次打印出每一个数字。
示例 1：
输入：matrix = [[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：matrix = [[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]
限制：
0 <= matrix.length <= 100
0 <= matrix[i].length <= 100
解题思路：
java
class Solution {
public int[] spiralOrder(int[][] matrix) {
if(matrix.length == 0) return new int[0];
int l = 0, r = matrix[0].length - 1, t = 0, b = matrix.length - 1, x = 0;
int[] res = new int[(r + 1) * (b + 1)];
while(true) {
for(int i = l; i <= r; i++) res[x++] = matrix[t][i]; // left to right.
if(++t > b) break;
for(int i = t; i <= b; i++) res[x++] = matrix[i][r]; // top to bottom.
if(l > --r) break;
for(int i = r; i >= l; i--) res[x++] = matrix[b][i]; // right to left.
if(t > --b) break;
for(int i = b; i >= t; i--) res[x++] = matrix[i][l]; // bottom to top.
if(++l > r) break;
}
return res;
}
}
python
class Solution(object):
def spiralOrder(self, matrix):
"""
:type matrix: List[List[int]]
:rtype: List[int]
"""
if not matrix :
return []
res = []
l = 0
r = len(matrix[0]) - 1
top = 0
low = len(matrix) - 1
while True:
for i in range(l,r+1):
res.append(matrix[top][i])
top = top + 1
if top>low:break
for j in range(top,low+1):
res.append(matrix[j][r])
r = r - 1
if r<l:break
for m in range(r,l-1,-1):
res.append(matrix[low][m])
low = low - 1
if low<top:break
for n in range(low,top-1,-1):
res.append(matrix[n][l])
l = l + 1
if l>r:
break
return res
