54 Spiral Matrix 螺旋矩阵
Description:
Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.
Example:
Example 1:
Input:
[
[ 1, 2, 3 ],
[ 4, 5, 6 ],
[ 7, 8, 9 ]
]
Output: [1,2,3,6,9,8,7,4,5]
Example 2:
Input:
[
[1, 2, 3, 4],
[5, 6, 7, 8],
[9,10,11,12]
]
Output: [1,2,3,4,8,12,11,10,9,5,6,7]
题目描述:
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 :
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]
思路:
按照螺旋的方法遍历整个矩阵
每一次遍历就更换方向
时间复杂度O(n), 空间复杂度O(n)
代码:
C++:
class Solution
{
public:
vector spiralOrder(vector>& matrix)
{
vector result;
if (matrix.empty() or matrix[0].empty()) return result;
int up = 0, down = matrix.size() - 1, left = 0, right = matrix[0].size() - 1;
while (true)
{
for (int i = left; i <= right; i++) result.push_back(matrix[up][i]);
if (++up > down) break;
for (int i = up; i <= down; i++) result.push_back(matrix[i][right]);
if (--right < left) break;
for (int i = right; i >= left; i--) result.push_back(matrix[down][i]);
if (--down < up) break;
for (int i = down; i >= up; i--) result.push_back(matrix[i][left]);
if (++left > right) break;
}
return result;
}
};
Java:
class Solution {
public List spiralOrder(int[][] matrix) {
if (matrix == null || matrix.length == 0 || matrix[0].length == 0) return new LinkedList<>();
List result = new LinkedList<>();
int up = 0, down = matrix.length - 1, left = 0, right = matrix[0].length - 1;
while (true) {
for (int i = left; i <= right; i++) result.add(matrix[up][i]);
if (++up > down) break;
for (int i = up; i <= down; i++) result.add(matrix[i][right]);
if (--right < left) break;
for (int i = right; i >= left; i--) result.add(matrix[down][i]);
if (--down < up) break;
for (int i = down; i >= up; i--) result.add(matrix[i][left]);
if (++left > right) break;
}
return result;
}
}
Python:
class Solution:
def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
if not matrix or not matrix[0]:
return []
result, up, down, left, right = [], 0, len(matrix) - 1, 0, len(matrix[0]) - 1
while True:
for i in range(left, right + 1):
result.append(matrix[up][i]);
up += 1
if up > down:
break
for i in range(up, down + 1):
result.append(matrix[i][right]);
right -= 1
if left > right:
break
for i in range(right, left - 1, - 1):
result.append(matrix[down][i]);
down -= 1
if up > down:
break
for i in range(down, up - 1, -1):
result.append(matrix[i][left]);
left += 1
if left > right:
break
return result