题目地址(54. 螺旋矩阵)

Posted 2022-05-31 潜行前行

题目地址(54. 螺旋矩阵)

https://leetcode.cn/problems/spiral-matrix/

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

关键点

• 边界判断

代码

• 语言支持：Java

Java Code:

class Solution 
    public List<Integer> spiralOrder(int[][] matrix) 
        List<Integer> order = new ArrayList<Integer>();
        if (matrix == null || matrix.length == 0 || matrix[0].length == 0) 
            return order;
        
        int rows = matrix.length, columns = matrix[0].length;
        int left = 0, right = columns - 1, top = 0, bottom = rows - 1;
        while (left <= right && top <= bottom) 
            for (int column = left; column <= right; column++) 
                order.add(matrix[top][column]);
            
            for (int row = top + 1; row <= bottom; row++) 
                order.add(matrix[row][right]);
            
            if (left < right && top < bottom) 
                for (int column = right - 1; column > left; column--) 
                    order.add(matrix[bottom][column]);
                
                for (int row = bottom; row > top; row--) 
                    order.add(matrix[row][left]);
                
            
            left++;
            right--;
            top++;
            bottom--;
        
        return order;