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 ] ]
跟Spiral Matrix一样的处理,甚至还简单一些,这个确定是方阵。分层,然后按照上右下左的顺序放入数组中。每个元素只访问一次,时间复杂度是O(n^2)。
1 public class Solution { 2 public int[][] generateMatrix(int n) { 3 int[][] res = new int[n][n]; 4 if (n < 0) return null; 5 int levelNum = n / 2; 6 int count = 1; 7 for (int level=0; level<levelNum; level++) { 8 for (int i=level; i<n-1-level; i++) { 9 res[level][i] = count++; 10 } 11 for (int i=level; i<n-1-level; i++) { 12 res[i][n-1-level] = count++; 13 } 14 for (int i=n-1-level; i>level; i--) { 15 res[n-1-level][i] = count++; 16 } 17 for (int i=n-1-level; i>level; i--) { 18 res[i][level] = count++; 19 } 20 } 21 if (n % 2 == 1) { 22 res[levelNum][levelNum] = count; 23 } 24 return res; 25 } 26 }
原文:http://www.cnblogs.com/EdwardLiu/p/4021422.html