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59. 螺旋矩阵 II

English Version

题目描述

给定一个正整数 n，生成一个包含 1 到 n² 所有元素，且元素按顺时针顺序螺旋排列的正方形矩阵。

示例:

输入: 3
输出:
[
 [ 1, 2, 3 ],
 [ 8, 9, 4 ],
 [ 7, 6, 5 ]
]

解法

Python3

class Solution:
    def generateMatrix(self, n: int) -> List[List[int]]:
        res = [[0] * n for _ in range(n)]
        num = 1
        m1, m2 = 0, n - 1
        while m1 < m2:
            for j in range(m1, m2):
                res[m1][j] = num
                num += 1
            for i in range(m1, m2):
                res[i][m2] = num
                num += 1
            for j in range(m2, m1, -1):
                res[m2][j] = num
                num += 1
            for i in range(m2, m1, -1):
                res[i][m1] = num
                num += 1
            m1 += 1
            m2 -= 1
        if m1 == m2:
            res[m1][m1] = num
        return res

Java

class Solution {
    public int[][] generateMatrix(int n) {
        int[][] res = new int[n][n];
        int num = 1;
        int m1 = 0, m2 = n - 1;
        while (m1 < m2) {
            for (int j = m1; j < m2; ++j) {
                res[m1][j] = num++;
            }
            for (int i = m1; i < m2; ++i) {
                res[i][m2] = num++;
            }
            for (int j = m2; j > m1; --j) {
                res[m2][j] = num++;
            }
            for (int i = m2; i > m1; --i) {
                res[i][m1] = num++;
            }
            ++m1;
            --m2;
        }
        if (m1 == m2) {
            res[m1][m1] = num;
        }

        return res;
    }
}

C++

class Solution {
public:
    vector<vector<int>> generateMatrix(int n) {
        vector<vector<int>> res(n, vector<int>(n, 0));
        int num = 1;
        int m1 = 0, m2 = n - 1;
        while (m1 < m2) {
            for (int j = m1; j < m2; ++j) {
                res[m1][j] = num++;
            }
            for (int i = m1; i < m2; ++i) {
                res[i][m2] = num++;
            }
            for (int j = m2; j > m1; --j) {
                res[m2][j] = num++;
            }
            for (int i = m2; i > m1; --i) {
                res[i][m1] = num++;
            }
            ++m1;
            --m2;
        }
        if (m1 == m2) {
            res[m1][m1] = num;
        }
        return res;
    }
};

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