LCP 54.螺旋矩阵

题目简述

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。

题解示例

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]

数据范围

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

标记难度

难度：中等
通过次数：158025
提交次数：335,777
通过率：47.1%

问题解析

本题主要考察矩阵的遍历，还涉及到递归、矩阵转置等知识。

最直观的方法就是按圈遍历矩阵，从外到里一圈一圈遍历，即先遍历矩阵的最外圈，然后递归地遍历其子矩阵的最外圈。要注意的是，单行或单列需要特殊处理。

python3代码

class Solution(object):
    def spiralOrder(self, matrix: List[List[int]]) -> List[int]: 
    	row, col = len(matrix), len(matrix[0])
        def getOutermostCircle(row_start, row_end, col_start, col_end): 
            if row_start > row_end or col_start > col_end:# 防错
                return []
            elif row_start == row_end: #单行
                circle = []
                for j in range(col_start, col_end+1): # 上边
                    circle.append(matrix[row_start][j])
                return circle
                # return matrix[row_start][col_start:col_end+1]
            elif col_start == col_end: #单列
                circle = []
                for i in range(row_start, row_end+1):
                    circle.append(matrix[i][col_start])
                return circle
            else: # 多行多列
                circle = []
                for j in range(col_start, col_end+1): # 上边
                    circle.append(matrix[row_start][j])
                    # circle.extend(matrix[row_start][col_start:col_end+1])
                for i in range(row_start+1, row_end): # 右边
                    circle.append(matrix[i][col_end])
                for j in range(col_end, col_start-1, -1): # 下边
                    circle.append(matrix[row_end][j])
                    # circle.extend(reversed(matrix[row_end][col_start:col_end+1]))
                for i in range(row_end-1, row_start, -1): # 左边
                    circle.append(matrix[i][col_start])
                return circle + getOutermostCircle(row_start+1, row_end-1, col_start+1, col_end-1) # 拼接

        return getOutermostCircle(0, row-1, 0, col-1)

C语言代码（官方）

int* spiralOrder(int** matrix, int matrixSize, int* matrixColSize, int* returnSize) {
    if (matrixSize == 0 || matrixColSize[0] == 0) {
        *returnSize = 0;
        return NULL;
    }

    int rows = matrixSize, columns = matrixColSize[0];
    int total = rows * columns;
    int* order = malloc(sizeof(int) * total);
    *returnSize = 0;

    int left = 0, right = columns - 1, top = 0, bottom = rows - 1;
    while (left <= right && top <= bottom) {
        for (int column = left; column <= right; column++) {
            order[(*returnSize)++] = matrix[top][column];
        }
        for (int row = top + 1; row <= bottom; row++) {
            order[(*returnSize)++] = matrix[row][right];
        }
        if (left < right && top < bottom) {
            for (int column = right - 1; column > left; column--) {
                order[(*returnSize)++] = matrix[bottom][column];
            }
            for (int row = bottom; row > top; row--) {
                order[(*returnSize)++] = matrix[row][left];
            }
        }
        left++;
        right--;
        top++;
        bottom--;
    }
    return order;
}

C++（官方）

class Solution {
public:
    vector<int> spiralOrder(vector<vector<int>>& matrix) {
        if (matrix.size() == 0 || matrix[0].size() == 0) {
            return {};
        }

        int rows = matrix.size(), columns = matrix[0].size();
        vector<int> order;
        int left = 0, right = columns - 1, top = 0, bottom = rows - 1;
        while (left <= right && top <= bottom) {
            for (int column = left; column <= right; column++) {
                order.push_back(matrix[top][column]);
            }
            for (int row = top + 1; row <= bottom; row++) {
                order.push_back(matrix[row][right]);
            }
            if (left < right && top < bottom) {
                for (int column = right - 1; column > left; column--) {
                    order.push_back(matrix[bottom][column]);
                }
                for (int row = bottom; row > top; row--) {
                    order.push_back(matrix[row][left]);
                }
            }
            left++;
            right--;
            top++;
            bottom--;
        }
        return order;
    }
};

大佬专属代码

def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
        res = []
        while matrix:
            res += matrix.pop(0)
            matrix = list(zip(*matrix))[::-1]
        return res

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熟练掌握矩阵、列表的各种函数或方法！！！

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