C++实现LeetCode(59.螺旋矩阵之二)

[LeetCode] 59. 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> generateMatrix(int n) {vector> res(n, vector(n)); 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)); 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; }};

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