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, 6, 5 ] ]
题解：以前做过的Spiral Matrix是给一个矩阵螺旋式的输出，这道题是给一个n，螺旋式的生成矩阵。和第一个的做法类似，定义四个变量
left:当前未填的子矩阵的最左边界
right:当前未填的子矩阵的最右边界
up:当前未填子矩阵的最上边界
down:当前未填子矩阵的最下边界
然后每次进行四个循环，做一个螺旋。
代码如下：
1 public class Solution { 2 public int[][] generateMatrix(int n) { 3 int[][] matrix = new int[n][n]; 4 int left = 0; 5 int right = n-1; 6 int up = 0; 7 int down = n-1; 8 int number = 0; 9 while(number <= n*n-1){ 10 for(int i = left;i <= right;i++){ 11 matrix[up][i] = number+1; 12 number++; 13 } 14 up++; 15 16 for(int i = up;i <= down;i++){ 17 matrix[i][right] = number+1; 18 number++; 19 } 20 right--; 21 22 for(int i = right;i >= left;i--){ 23 matrix[down][i] = number+1; 24 number++; 25 } 26 down--; 27 28 for(int i = down;i >= up;i--){ 29 matrix[i][left] = number+1; 30 number++; 31 } 32 left++; 33 } 34 return matrix; 35 } 36 }