https://leetcode.com/problems/spiral-matrix/ 题意分析: 输入一个m×n的数字矩阵。将这个矩阵按照螺旋方向输成一列。比如: [ [ 1, 2, 3 ], [ 4, 5, 6 ], [ 7, 8, 9 ] ] 输出[1,2,3,6,9,8,7,4,5] 题目思路: 这道题目只需要模拟这个过程。分四步,第一步行左到右,第二步从上到下...
自己本来在想有没有一些不一样的算法,然后发现直接右下左上顺序做就好了,看到一个博客的分类做得不错,重点是要看边界情况,直接看代码。 AC代码(Python) 1classSolution(object):2defspiralOrder(self, matrix):3"""4:type matrix: List[List[int]]5:rtype: List[int]6"""7ifmatrix ==[]:8return[]9...
题目地址:https://leetcode.com/problems/spiral-matrix-ii/description/ 题目描述 Given an integer n, generate a square matrix filled with elements from 1 to n2 in spiral order. For example, Given n = 3, You should return the following matrix: [ [ 1, 2, 3 ], [ 8, 9, 4 ], [ 7,...
输入: matrix = [ [1, 2, 3], [4, 5, 6], [7, 8, 9] ] 输出: [1, 2, 3, 6, 9, 8, 7, 4, 5]
[Leetcode][python]Spiral Matrix/Spiral Matrix II/螺旋矩阵/螺旋矩阵 II,SpiralMatrix题目大意将一个矩阵中的内容螺旋输出。注意点:矩阵不一定是正方形例子:输入:matrix=[[1,2,3],[4,5,6],[7,8,9]]输出:[1,2,3,6,9,8,7,4,5]解题思路看代码就可以理解代码classSolution(
Can you solve this real interview question? Spiral Matrix - Given an m x n matrix, return all elements of the matrix in spiral order. Example 1: [https://assets.leetcode.com/uploads/2020/11/13/spiral1.jpg] Input: matrix = [[1,2,3],[4,5,6],[7,8,9]]
/** direction 0 代表向右, 1 代表向下, 2 代表向左, 3 代表向上*/publicint[][]generateMatrix(intn){int[][]ans=newint[n][n];intstart_x=0,start_y=0,direction=0,top_border=-1,// 上边界right_border=n,// 右边界bottom_border=n,// 下边界left_border=-1;// 左边界intcount=1;while...
https://leetcode.com/problems/spiral-matrix/discuss/20571/1-liner-in-Python-%2B-Ruby Solutions 代码语言:javascript 代码运行次数:0 运行 AI代码解释 1publicList<Integer>spiralOrder(int[][]matrix){2List<Integer>res=newArrayList<>();34if(matrix==null||matrix.length==0||matrix[0].length==0){...
因为如果 matrix.length==0 判断为true,则后面的 matrix[0].length==0 不会再判断,即返回空数组;但是matrix[0].length==0 在前时,如果输入数组为空,matrix[0] 会报错因为matrix并没有0号索引。python3:class Solution: def spiralOrder(self, matrix: List[List[int]]) -> List[int]: if len(matrix...
(matrix[i][right])foriinrange(right,left,-1):result.append(matrix[bottom][i])foriinrange(bottom,top,-1):result.append(matrix[i][left])left+=1right-=1top+=1bottom-=1ifleft==rightandtop==bottom:result.append(matrix[top][left])elifleft==right:foriinrange(top,bottom+1):result....