54. 螺旋矩阵

题目描述

给你一个 m 行 n 列的矩阵 matrix ，请按照 顺时针螺旋顺序 ，返回矩阵中的所有元素。

示例 1：

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

示例 2：

输入：matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12]]
输出：[1,2,3,4,8,12,11,10,9,5,6,7]

提示：

• m == matrix.length
• n == matrix[i].length
• 1 <= m, n <= 10
• -100 <= matrix[i][j] <= 100

方法一：模拟

我们用 $i$ 和 $j$ 分别表示当前访问到的元素的行和列，用 $k$ 表示当前的方向，用数组或哈希表 $vis$ 记录每个元素是否被访问过。每次我们访问到一个元素后，将其标记为已访问，然后按照当前的方向前进一步，如果前进一步后发现越界或者已经访问过，则改变方向继续前进，直到遍历完整个矩阵。

时间复杂度 $O(m\times n)$，空间复杂度 $O(m\times n)$。其中 $m$ 和 $n$ 分别是矩阵的行数和列数。

对于访问过的元素，我们也可以将其值加上一个常数 $300$，这样就不需要额外的 $vis$ 数组或哈希表来记录是否访问过了，从而将空间复杂度降低到 $O(1)$。

Java

class Solution {
    public List<Integer> spiralOrder(int[][] matrix) {
        int m = matrix.length, n = matrix[0].length;
        int[] dirs = {0, 1, 0, -1, 0};
        int i = 0, j = 0, k = 0;
        List<Integer> ans = new ArrayList<>();
        boolean[][] vis = new boolean[m][n];
        for (int h = m * n; h > 0; --h) {
            ans.add(matrix[i][j]);
            vis[i][j] = true;
            int x = i + dirs[k], y = j + dirs[k + 1];
            if (x < 0 || x >= m || y < 0 || y >= n || vis[x][y]) {
                k = (k + 1) % 4;
            }
            i += dirs[k];
            j += dirs[k + 1];
        }
        return ans;
    }
}

C++

class Solution {
public:
    vector<int> spiralOrder(vector<vector<int>>& matrix) {
        int m = matrix.size(), n = matrix[0].size();
        int dirs[5] = {0, 1, 0, -1, 0};
        int i = 0, j = 0, k = 0;
        vector<int> ans;
        bool vis[m][n];
        memset(vis, false, sizeof(vis));
        for (int h = m * n; h; --h) {
            ans.push_back(matrix[i][j]);
            vis[i][j] = true;
            int x = i + dirs[k], y = j + dirs[k + 1];
            if (x < 0 || x >= m || y < 0 || y >= n || vis[x][y]) {
                k = (k + 1) % 4;
            }
            i += dirs[k];
            j += dirs[k + 1];
        }
        return ans;
    }
};

TypeScript

function spiralOrder(matrix: number[][]): number[] {
    const m = matrix.length;
    const n = matrix[0].length;
    const ans: number[] = [];
    const vis = new Array(m).fill(0).map(() => new Array(n).fill(false));
    const dirs = [0, 1, 0, -1, 0];
    for (let h = m * n, i = 0, j = 0, k = 0; h > 0; --h) {
        ans.push(matrix[i][j]);
        vis[i][j] = true;
        const x = i + dirs[k];
        const y = j + dirs[k + 1];
        if (x < 0 || x >= m || y < 0 || y >= n || vis[x][y]) {
            k = (k + 1) % 4;
        }
        i += dirs[k];
        j += dirs[k + 1];
    }
    return ans;
}

Python

class Solution:
    def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
        m, n = len(matrix), len(matrix[0])
        dirs = (0, 1, 0, -1, 0)
        i = j = k = 0
        ans = []
        vis = set()
        for _ in range(m * n):
            ans.append(matrix[i][j])
            vis.add((i, j))
            x, y = i + dirs[k], j + dirs[k + 1]
            if not 0 <= x < m or not 0 <= y < n or (x, y) in vis:
                k = (k + 1) % 4
            i = i + dirs[k]
            j = j + dirs[k + 1]
        return ans

方法二：逐层模拟

我们也可以从外往里一圈一圈遍历并存储矩阵元素。

时间复杂度 $O(m\times n)$，空间复杂度 $O(1)$。其中 $m$ 和 $n$ 分别是矩阵的行数和列数。

Java

class Solution {
    public List<Integer> spiralOrder(int[][] matrix) {
        int m = matrix.length, n = matrix[0].length;
        int[] dirs = {0, 1, 0, -1, 0};
        List<Integer> ans = new ArrayList<>();
        for (int h = m * n, i = 0, j = 0, k = 0; h > 0; --h) {
            ans.add(matrix[i][j]);
            matrix[i][j] += 300;
            int x = i + dirs[k], y = j + dirs[k + 1];
            if (x < 0 || x >= m || y < 0 || y >= n || matrix[x][y] > 100) {
                k = (k + 1) % 4;
            }
            i += dirs[k];
            j += dirs[k + 1];
        }
        // for (int i = 0; i < m; ++i) {
        //     for (int j = 0; j < n; ++j) {
        //         matrix[i][j] -= 300;
        //     }
        // }
        return ans;
    }
}

C++

class Solution {
public:
    vector<int> spiralOrder(vector<vector<int>>& matrix) {
        int m = matrix.size(), n = matrix[0].size();
        int dirs[5] = {0, 1, 0, -1, 0};
        vector<int> ans;
        for (int h = m * n, i = 0, j = 0, k = 0; h; --h) {
            ans.push_back(matrix[i][j]);
            matrix[i][j] += 300;
            int x = i + dirs[k], y = j + dirs[k + 1];
            if (x < 0 || x >= m || y < 0 || y >= n || matrix[x][y] > 100) {
                k = (k + 1) % 4;
            }
            i += dirs[k];
            j += dirs[k + 1];
        }
        // for (int i = 0; i < m; ++i) {
        //     for (int j = 0; j < n; ++j) {
        //         matrix[i][j] -= 300;
        //     }
        // }
        return ans;
    }
};

TypeScript

function spiralOrder(matrix: number[][]): number[] {
    const m = matrix.length;
    const n = matrix[0].length;
    const ans: number[] = [];
    const dirs = [0, 1, 0, -1, 0];
    for (let h = m * n, i = 0, j = 0, k = 0; h > 0; --h) {
        ans.push(matrix[i][j]);
        matrix[i][j] += 300;
        const x = i + dirs[k];
        const y = j + dirs[k + 1];
        if (x < 0 || x >= m || y < 0 || y >= n || matrix[x][y] > 100) {
            k = (k + 1) % 4;
        }
        i += dirs[k];
        j += dirs[k + 1];
    }
    // for (let i = 0; i < m; ++i) {
    //     for (let j = 0; j < n; ++j) {
    //         matrix[i][j] -= 300;
    //     }
    // }
    return ans;
}

Python

class Solution:
    def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
        m, n = len(matrix), len(matrix[0])
        dirs = (0, 1, 0, -1, 0)
        i = j = k = 0
        ans = []
        for _ in range(m * n):
            ans.append(matrix[i][j])
            matrix[i][j] += 300
            x, y = i + dirs[k], j + dirs[k + 1]
            if not 0 <= x < m or not 0 <= y < n or matrix[x][y] > 100:
                k = (k + 1) % 4
            i = i + dirs[k]
            j = j + dirs[k + 1]
        # for i in range(m):
        #     for j in range(n):
        #         matrix[i][j] -= 300
        return ans

方法三

Java

class Solution {
    public List<Integer> spiralOrder(int[][] matrix) {
        int m = matrix.length, n = matrix[0].length;
        int x1 = 0, y1 = 0, x2 = m - 1, y2 = n - 1;
        List<Integer> ans = new ArrayList<>();
        while (x1 <= x2 && y1 <= y2) {
            for (int j = y1; j <= y2; ++j) {
                ans.add(matrix[x1][j]);
            }
            for (int i = x1 + 1; i <= x2; ++i) {
                ans.add(matrix[i][y2]);
            }
            if (x1 < x2 && y1 < y2) {
                for (int j = y2 - 1; j >= y1; --j) {
                    ans.add(matrix[x2][j]);
                }
                for (int i = x2 - 1; i > x1; --i) {
                    ans.add(matrix[i][y1]);
                }
            }
            ++x1;
            ++y1;
            --x2;
            --y2;
        }
        return ans;
    }
}

C++

class Solution {
public:
    vector<int> spiralOrder(vector<vector<int>>& matrix) {
        int m = matrix.size(), n = matrix[0].size();
        int x1 = 0, y1 = 0, x2 = m - 1, y2 = n - 1;
        vector<int> ans;
        while (x1 <= x2 && y1 <= y2) {
            for (int j = y1; j <= y2; ++j) {
                ans.push_back(matrix[x1][j]);
            }
            for (int i = x1 + 1; i <= x2; ++i) {
                ans.push_back(matrix[i][y2]);
            }
            if (x1 < x2 && y1 < y2) {
                for (int j = y2 - 1; j >= y1; --j) {
                    ans.push_back(matrix[x2][j]);
                }
                for (int i = x2 - 1; i > x1; --i) {
                    ans.push_back(matrix[i][y1]);
                }
            }
            ++x1, ++y1;
            --x2, --y2;
        }
        return ans;
    }
};

TypeScript

function spiralOrder(matrix: number[][]): number[] {
    const m = matrix.length;
    const n = matrix[0].length;
    let x1 = 0;
    let y1 = 0;
    let x2 = m - 1;
    let y2 = n - 1;
    const ans: number[] = [];
    while (x1 <= x2 && y1 <= y2) {
        for (let j = y1; j <= y2; ++j) {
            ans.push(matrix[x1][j]);
        }
        for (let i = x1 + 1; i <= x2; ++i) {
            ans.push(matrix[i][y2]);
        }
        if (x1 < x2 && y1 < y2) {
            for (let j = y2 - 1; j >= y1; --j) {
                ans.push(matrix[x2][j]);
            }
            for (let i = x2 - 1; i > x1; --i) {
                ans.push(matrix[i][y1]);
            }
        }
        ++x1;
        ++y1;
        --x2;
        --y2;
    }
    return ans;
}

Python

class Solution:
    def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
        m, n = len(matrix), len(matrix[0])
        x1, y1, x2, y2 = 0, 0, m - 1, n - 1
        ans = []
        while x1 <= x2 and y1 <= y2:
            for j in range(y1, y2 + 1):
                ans.append(matrix[x1][j])
            for i in range(x1 + 1, x2 + 1):
                ans.append(matrix[i][y2])
            if x1 < x2 and y1 < y2:
                for j in range(y2 - 1, y1 - 1, -1):
                    ans.append(matrix[x2][j])
                for i in range(x2 - 1, x1, -1):
                    ans.append(matrix[i][y1])
            x1, y1 = x1 + 1, y1 + 1
            x2, y2 = x2 - 1, y2 - 1
        return ans