we traverse the matrix as the spiral order. Each loop we go through the four edges of the periphery of the remaining matrix. The number of the periphery is n/2 and the number of elements in each edge of the peripehry is n-1-2*i
Thus we can have the code:
public class Solution {
public int[][] generateMatrix(int n) {
if(n < 0) n = 0 - n;
int[][] matrix = new int[n][n];
if(n == 0) return matrix;
int count = 0;
for(int i = 0; i < n/2; i++){
int num = n-1-i*2;
for(int j = 0; j < num; j++){
matrix[i][i+j] = count+j+1;
matrix[i+j][i+num] = count+num+j+1;
matrix[i+num][i+num-j] = count+2*num+j+1;
matrix[i+num-j][i] = count+3*num+j+1;
}
count += 4*num;
}
if(n%2 == 1) matrix[n/2][n/2] = n*n;
return matrix;
}
}
Jan 18 - Spiral Matrix II; 2D Array;
原文：http://www.cnblogs.com/5683yue/p/5141862.html