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

解法

从外往里一圈一圈遍历并存储矩阵元素即可。

Python3

class Solution:
    def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
        def add(i1, j1, i2, j2):
            if i1 == i2:
                return [matrix[i1][j] for j in range(j1, j2 + 1)]
            if j1 == j2:
                return [matrix[i][j1] for i in range(i1, i2 + 1)]
            return [matrix[i1][j] for j in range(j1, j2)] + [matrix[i][j2] for i in range(i1, i2)] + [matrix[i2][j] for j in range(j2, j1, -1)] + [matrix[i][j1] for i in range(i2, i1, -1)]
        m, n = len(matrix), len(matrix[0])
        i1, j1, i2, j2 = 0, 0, m - 1, n - 1
        res = []
        while i1 <= i2 and j1 <= j2:
            res += add(i1, j1, i2, j2)
            i1, j1, i2, j2 = i1 + 1, j1 + 1, i2 - 1, j2 - 1
        return res

Java

class Solution {
    private List<Integer> res;

    public List<Integer> spiralOrder(int[][] matrix) {
        int m = matrix.length, n = matrix[0].length;
        res = new ArrayList<>();
        int i1 = 0, i2 = m - 1;
        int j1 = 0, j2 = n - 1;
        while (i1 <= i2 && j1 <= j2) {
            add(matrix, i1++, j1++, i2--, j2--);
        }
        return res;
    }

    private void add(int[][] matrix, int i1, int j1, int i2, int j2) {
        if (i1 == i2) {
            for (int j = j1; j <= j2; ++j) {
                res.add(matrix[i1][j]);
            }
            return;
        }
        if (j1 == j2) {
            for (int i = i1; i <= i2; ++i) {
                res.add(matrix[i][j1]);
            }
            return;
        }
        for (int j = j1; j < j2; ++j) {
            res.add(matrix[i1][j]);
        }
        for (int i = i1; i < i2; ++i) {
            res.add(matrix[i][j2]);
        }
        for (int j = j2; j > j1; --j) {
            res.add(matrix[i2][j]);
        }
        for (int i = i2; i > i1; --i) {
            res.add(matrix[i][j1]);
        }
    }
}

JavaScript

/**
 * @param {number[][]} matrix
 * @return {number[]}
 */
var spiralOrder = function (matrix) {
    let m = matrix.length;
    if (m === 0) return [];
    let res = [];
    let top = 0, bottom = m - 1, left = 0, right = matrix[0].length - 1;
    while (left < right && bottom > top) {
        for (let i = left; i < right; i++) res.push(matrix[top][i]);
        for (let i = top; i < bottom; i++) res.push(matrix[i][right]);
        for (let i = right; i > left; i--) res.push(matrix[bottom][i]);
        for (let i = bottom; i > top; i--) res.push(matrix[i][left]);
        top++;
        bottom--;
        left++;
        right--;
    }
    if (left === right) {
        for (i = top; i <= bottom; i++) res.push(matrix[i][left]);
    } else if (top === bottom) {
        for (i = left; i <= right; i++) res.push(matrix[top][i]);
    }
    return res;
};

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