29. 顺时针打印矩阵
1. 描述
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。
2. 例子
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]
3. 限制
- 0 <= matrix.length <= 100
- 0 <= matrix[i].length <= 100
4. 题解
n 是 i_matrix 中的数字数
时间复杂度: O(n)
空间复杂度: O(n)
class Solution
{
public:
// vector<int> spiralOrder(vector<vector<int>>& matrix)
vector<int> spiralOrder(vector<vector<int>> &i_matrix)
{
if(i_matrix.empty() || i_matrix.front().empty()) return vector<int>{};
int leftBorder = 0, rightBorder = i_matrix.front().size() - 1, upBorder = 1, downBorder = i_matrix.size() - 1;
int direction = (i_matrix.front().size() > 1 ? 0 : 1); // 0 1 2 3: right, down, left, up
vector<int> result(i_matrix.size() * i_matrix.front().size());
for(int i = 0, row = 0, column = 0; i < i_matrix.size() * i_matrix.front().size(); i++)
{
result[i] = i_matrix[row][column];
if(direction == 0)
{
column++;
if(column == rightBorder)
{
direction = (direction + 1) % 4;
rightBorder--;
}
}
else if(direction == 1)
{
row++;
if(row == downBorder)
{
direction = (direction + 1) % 4;
downBorder--;
}
}
else if(direction == 2)
{
column--;
if(column == leftBorder)
{
direction = (direction + 1) % 4;
leftBorder++;
}
}
else if(direction == 3)
{
row--;
if(row == upBorder)
{
direction = (direction + 1) % 4;
upBorder++;
}
}
}
return result;
}
};