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]; 
                    }
                }
                
            }
            
        }        
    }
}