[LeetCode] Spiral Matrix II 螺旋矩阵之二

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，参见代码如下：

解法一：

class Solution {

public:

    vector<vector<int>> generateMatrix(int n) {

        vector<vector<int>> res(n, vector<int>(n));

        int val = , p = n;

        for (int i = ; i < n / ; ++i, p -= ) {

            for (int col = i; col < i + p; ++col)

                res[i][col] = val++;

            for (int row = i + ; row < i + p; ++row)

                res[row][i + p - ] = val++;

            for (int col = i + p - ; col >= i; --col)

                res[i + p - ][col] = val++;

            for (int row = i + p - ; row > i; --row)

                res[row][i] = val++;

        }

        if (n %  != ) res[n / ][n / ] = 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));

        int up = , down = n - , left = , right = n - , val = ;

        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;

    }

};