Given a positive integer n, generate a square matrix filled with elements from 1 to n2 in spiral order.
Example:
Input: 3 Output: [ [ 1, 2, 3 ], [ 8, 9, 4 ], [ 7, 6, 5 ] ]
此题跟之前那道 Spiral Matrix 本质上没什么区别，就相当于个类似逆运算的过程，这道题是要按螺旋的顺序来填数，由于给定矩形是个正方形，我们计算环数时用n / 2来计算，若n为奇数时，此时最中间的那个点没有被算在环数里，所以最后需要单独赋值，还是下标转换问题是难点，参考之前 Spiral Matrix 的讲解来转换下标吧，代码如下：
解法一：
class Solution { public: vector<vector<int> > generateMatrix(int n) { vector<vector<int> > res(n, vector<int>(n, 0)); int val = 1, p = n; for (int i = 0; i < n / 2; ++i, p -= 2) { for (int col = i; col < i + p; ++col) res[i][col] = val++; for (int row = i + 1; row < i + p; ++row) res[row][i + p - 1] = val++; for (int col = i + p - 2; col >= i; --col) res[i + p - 1][col] = val++; for (int row = i + p - 2; row > i; --row) res[row][i] = val++; } if (n % 2 != 0) res[n / 2][n / 2] = val; return res; } };
当然我们也可以使用下面这种简化了坐标转换的方法，博主个人还是比较推崇下面这种解法，不容易出错，而且好理解，参见代码如下：
解法二：
class Solution { public: vector<vector<int>> generateMatrix(int n) { vector<vector<int>> res(n, vector<int>(n, 0)); int up = 0, down = n - 1, left = 0, right = n - 1, val = 1; while (true) { for (int j = left; j <= right; ++j) res[up][j] = val++; if (++up > down) break; for (int i = up; i <= down; ++i) res[i][right] = val++; if (--right < left) break; for (int j = right; j >= left; --j) res[down][j] = val++; if (--down < up) break; for (int i = down; i >= up; --i) res[i][left] = val++; if (++left > right) break; } return res; } };
类似题目：
参考资料：
https://leetcode.com/problems/spiral-matrix-ii/
