Print the matrix clockwise
Enter a matrix and print out each number in clockwise order from the outside in.
Example 1:
Input: matrix = [[1, 2, 3], [4 and 6], [7,8,9]] output:,2,3,6,9,8,7,4,5 [1] example 2:
Input: matrix = [[1, 2, 3, 4], [5,6,7,8], [9,10,11,12]] output:,2,3,4,8,12,11,10,9,5,6,7 [1]
Limitations:
0 <= matrix.length <= 100
0 <= matrix[i].length <= 100
Answer:
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;
}
returnres; }}Copy the code
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
Copy the code