Given an integern, generate a square matrix filled with elements from 1 ton2 in spiral order. For example, Givenn=3, You should return the following matrix: [ [ 1, 2, 3 ], [ 8, 9, 4 ], [ 7, 6, 5 ] ] 1publicclass
59. Spiral Matrix II(螺旋矩阵II) 原题链接:https://leetcode.com/problems/spiral-matrix-ii/ 思路和上篇的螺旋矩阵一样,用两个一维数组控制方向, 用一个Boolean数组判断是否访问过,然后更新 i , j的值。 螺旋矩阵I详细解释 AC 1ms 100% Java: ......
Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order. For example, Given the following matrix: [ [ 1, 2, 3 ], [ 4, 5, 6 ], [ 7, 8, 9 ] ] You should return [1,2,3,6,9,8,7,4,5]. Solution 1: 使用递归,一次扫描...
Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order. For example, Given the following matrix: [ [ 1, 2, 3 ], [ 4, 5, 6 ], [ 7, 8, 9 ] ] You should return[1,2,3,6,9,8,7,4,5]. 顺序添加法 复杂度 时间O(NM) ...
题目1. 输出螺旋矩阵https://leetcode.com/problems/spiral-matrix/ 题目描述:顺时针方向输出螺旋矩阵,逐步螺旋进中心, 将元素按此顺序输出 题解: 略 java顺时针螺旋遍历M*N的矩阵 java顺时针螺旋遍历M*N的矩阵给定一个包含mxn个元素的矩阵(m行,n列),请按照顺时针螺旋顺序,返回矩阵中的所有元素。示例1:示例2...
class Solution { public int[][] generateMatrix(int n) { int[][] arr = new int[n][n]; //打印几圈 int loops = (n+1)/2; int num = 1; for(int i=0; i<loops; i++){ //每一圈打印的起点(i,i) //左到右; i-0 = n-1-y -> y=n-1-i ...
public class Solution { public int[][] generateMatrix(int n) { int[][] result = new int[n][n]; int layer; int k; for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { layer = layer(i, j, n); // 当前坐标外有几层 ...
There is this awesome one line solution from this guy which is pretty insane. 代码语言:javascript 代码运行次数:0 运行 AI代码解释 defspiralOrder(self,matrix):returnmatrix andlist(matrix.pop(0))+self.spiralOrder(zip(*matrix)[::-1])
matrix[startx][y] = num++; } // 右边的列,从上到下 for (int x = startx + 1; x <= endx; x++) { matrix[x][endy] = num++; } // 如果行或列遍历完,则退出循环 if (startx == endx || starty == endy) { break;
class Solution { public: vector<int> spiralOrder(vector<vector<int>>& matrix) { vector<int> res; if(matrix.size() == 0) return {}; int colbegin = 0,colend = matrix[0].size()-1; int rowbegin = 0,rowend = matrix.size()-1; while(rowbegin <= rowend && colbegin <= colend...