Spiral Matrix II

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 ]
]

这道题接着上面那道题，是一个逆过程，只要稍微改一下就好了，因为上面一题螺旋式的遍历了整个数组，只要在遍历过程中将数字填进去就好了

 1 import java.util.ArrayList;
 2 import java.util.Arrays;
 3 import java.util.List;
 4 
 5 
 6 
 7 public class Solution {
 8 //    List<Integer> result = new ArrayList<Integer>();
 9     int result[][];
10     int count = 1;
11     public int[][] generateMatrix(int n) {
12         
13         result = new int[n][n];
14         if(0 == n)
15             return result;
16         spiralOrder(result);
17         return result;
18     }
19     
20     public void spiralOrder(int[][] matrix) {
21         
22         
23         if(matrix.length == 1 && matrix[0].length == 1)    //只有一个元素直接添加
24         {
25             matrix[0][0] = count++;
26         }
27         else if(matrix[0].length == 1){                        //竖条
28             for(int i = 0; i < matrix.length; i++){
29                 matrix[i][0] = count++;
30             }   
31         }
32         else if(matrix.length == 1){                    //横条
33             for(int i = 0; i < matrix[0].length; i++){
34                 matrix[0][i] = count++;
35             }
36         }
37         else {
38             for(int i = 0; i < matrix[0].length; i++){    //添加第一排
39                 matrix[0][i] = count++;
40             }
41             for(int i = 1; i < matrix.length; i++){        //添加最后一竖
42                 matrix[i][matrix[0].length - 1] = count++;
43             }
44             for(int i = matrix[0].length - 2; i >= 0; i--){    //添加最后一排
45                 matrix[matrix.length - 1][i] = count++;
46             }
47             for(int i = matrix.length - 2; i >= 1;i--){        //添加第一排
48                 matrix[i][0] = count++;
49             }
50             if(matrix.length - 2 != 0 && matrix[0].length - 2 != 0){
51                 int next[][] = new int[matrix.length - 2][matrix[0].length - 2];
52                 spiralOrder(next);                                //递归求解下一个矩阵的值
53                 for(int i = 1; i < matrix.length - 1; i++){
54                     for(int j = 1; j < matrix[0].length - 1;j++){
55                         matrix[i][j] = next[i - 1][j - 1]; 
56                     }
57                 }
58                 
59             }
60             
61         }        
62     }
63 }