Given a positive integer n, generate a square matrix filled with elements from 1 to n2 in spiral order.

Example:

Input: 3Output:[ [ 1, 2, 3 ], [ 8, 9, 4 ], [ 7, 6, 5 ]]

 

此题跟之前那道 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 的讲解来转换下标吧，代码如下：

 

解法一：

class Solution {public:    vector
     
      
        > generateMatrix(int n) {        vector
       
        
          > res(n, vector
         
          (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
     
      
       > generateMatrix(int n) {        vector
       
        
         > res(n, vector
         
          (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：