59.md

中规中矩】59. 螺旋矩阵 II (旋转生成1~n^2)

解题思路

考验编程技巧，下面的写法相对比较好懂好记忆。基本思路就是定义四个边界的范围，围绕矩阵四周上下左右逐行逐列输出，同时更新上下左右的行列坐标的值即可。python和C++两种语言实现。

代码

class Solution {
public:
    vector<vector<int>> generateMatrix(int n) {
        vector<vector<int>>  res(n, vector<int>(n));
        int count = 1;
        int top = 0;
        int bottom = n - 1;
        int left = 0;
        int right = n - 1;

        while (true) {
            // fill top row
            for (int i = left; i <= right; i++) {
                res[top][i] = count++;
            }
            top++;
            if (top > bottom) {
                break;
            }

            // fill right col
            for (int j = top; j <= bottom; j++) {
                res[j][right] = count++;
            }
            right--;
            if (right < left) {
                break;
            }

            // fill bottom row
            for (int k = right; k >= left; k--) {
                res[bottom][k] = count++;
            }
            bottom--;
            if (bottom < top) {
                break;
            }

            // fill left col
            for (int l = bottom; l >= top; l--) {
                res[l][left] = count++;
            }
            left++;
            if (left > right) {
                break;
            }
        }

        return res;
    }
};

class Solution(object):
    def generateMatrix(self, n):
        res = [[0] * n for _ in range(n)]
        count = 1
        top = 0
        bottom = n - 1
        left = 0
        right = n - 1

        while (True):
            for i in range(left, right + 1):
                res[top][i] = count
                count += 1
            top += 1
            if (top > bottom):
                break
            
            for j in range(top, bottom + 1):
                res[j][right] = count
                count += 1
            right -= 1
            if (right < left):
                break
            
            for k in range(right, left - 1, -1):
                res[bottom][k] = count
                count += 1
            bottom -= 1
            if (bottom < top):
                break
            
            for l in range(bottom, top - 1, -1):
                res[l][left] = count
                count += 1
            left += 1
            if (left > right):
                break
            

        return res

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