Spiral Matrix II

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

 import java.util.ArrayList;
 import java.util.Arrays;
 import java.util.List;

 public class Solution {
 //    List<Integer> result = new ArrayList<Integer>();
     int result[][];
     int count = 1;
     public int[][] generateMatrix(int n) {

         result = new int[n][n];
         if(0 == n)
             return result;
         spiralOrder(result);
         return result;
     }

     public void spiralOrder(int[][] matrix) {

         if(matrix.length == 1 && matrix[0].length == 1)    //只有一个元素直接添加
         {
             matrix[0][0] = count++;
         }
         else if(matrix[0].length == 1){                        //竖条
             for(int i = 0; i < matrix.length; i++){
                 matrix[i][0] = count++;
             }
         }
         else if(matrix.length == 1){                    //横条
             for(int i = 0; i < matrix[0].length; i++){
                 matrix[0][i] = count++;
             }
         }
         else {
             for(int i = 0; i < matrix[0].length; i++){    //添加第一排
                 matrix[0][i] = count++;
             }
             for(int i = 1; i < matrix.length; i++){        //添加最后一竖
                 matrix[i][matrix[0].length - 1] = count++;
             }
             for(int i = matrix[0].length - 2; i >= 0; i--){    //添加最后一排
                 matrix[matrix.length - 1][i] = count++;
             }
             for(int i = matrix.length - 2; i >= 1;i--){        //添加第一排
                 matrix[i][0] = count++;
             }
             if(matrix.length - 2 != 0 && matrix[0].length - 2 != 0){
                 int next[][] = new int[matrix.length - 2][matrix[0].length - 2];
                 spiralOrder(next);                                //递归求解下一个矩阵的值
                 for(int i = 1; i < matrix.length - 1; i++){
                     for(int j = 1; j < matrix[0].length - 1;j++){
                         matrix[i][j] = next[i - 1][j - 1];
                     }
                 }

             }

         }
     }
 }