Spiral Matrix II

Simple question. Be careful of the traversal direction, and length change after array is traversed horizontally.

 public class Solution {  
   public int[][] generateMatrix(int n) {  
     int[][] res = new int[n][n];  
     if(n<1) return res;      
     int dir = 0;// 0 means direction is from left to right, 1 means from up to down, 2 right to up,  
             // 3 down to up  
     int length = n;  
     int val = 1, i = 0, j = 0;  
     while(val <= Math.pow(n,2)){  
       int steps = 0;  
       if(dir == 0){// i is fixed, j should increase  
         for(;steps < length && val <= Math.pow(n,2);steps++){ res[i][j+steps] = val; val++;}  
         dir = (++dir)%4;  
         length --;  
         i++;  
         j = j + steps - 1;  
       }else{  
         if(dir == 1){// j is fixed, i should increase  
           for(;steps < length && val <= Math.pow(n,2);steps++){ res[i+steps][j] = val; val++;}  
           dir = (++dir)%4;  
           j--;  
           i = i + steps - 1;  
         }  
         else{  
           if(dir == 2){// i is fixed, j should decrease  
             for(;steps < length && val <= Math.pow(n,2);steps ++){ res[i][j-steps] = val; val++;}  
             dir = (++dir)%4;  
             length --;  
             i --;  
             j = j - (steps - 1);  
           }  
           else{  
              if(dir == 3){// j is fixed, i should decrease  
               for(;steps < length && val <= Math.pow(n,2);steps ++){ res[i - steps][j] = val; val++;}  
               dir = (++dir)%4;  
               j ++;  
               i = i - (steps - 1);  
              }             
           }  
         }  
       }  
     }  
     return res;  
   }  
 }  

UPDATE:

Recently I was preparing for one interview, and happened to work on this question again. Below is another implementation of my algorithm, which is much cleaner:

 public class Solution {  
   public int[][] generateMatrix(int n) {  
     //if(n <= 0) return null;  
     int[][] res = new int[n][n];  
     int m = 1, i = 0, j = 0;  
     int end = n*n;//Math.pow(n,2);  
     while(m <= end){  
       for(j = i; (j <= n - 1 - i) && ( m <= end) ; j++){ res[i][j] = m ; m ++;} j--;  
       for(i++; i <= j && m <= end; i++){ res[i][j] = m ; m ++;} i--;  
       for(j--; (j >= n-1-i) && (m <= end); j--){ res[i][j] = m ; m ++;} j++;  
       for(i--; i >= j+1 && m <= end; i--){ res[i][j] = m ; m ++;} i++;  
     }  
     return res;  
   }  
 }