54 59. Spiral Matrix I II

Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.

For example,
Given the following matrix:

[ [ 1, 2, 3 ], [ 4, 5, 6 ], [ 7, 8, 9 ]]

You should return [1,2,3,6,9,8,7,4,5].

就是找规律，四个循环，注意边界！！

public static List<Integer> spiralOrder(int[][] matrix){List<Integer> retlist=new ArrayList<>();int m=matrix.length-1;if(m<0)return retlist;int n=matrix[0].length-1;int rowtimes=(m+2)/2;int coltimes=(n+2)/2;for(int cnt=0;cnt<Math.max(rowtimes, coltimes);cnt++){if(cnt<rowtimes){for(int j=cnt;j<=n-cnt;j++)if(cnt<=m)retlist.add(matrix[cnt][j]);}if(cnt<coltimes){for(int i=cnt+1;i<=m-cnt-1;i++)if(n-cnt>=0)retlist.add(matrix[i][n-cnt]);}if(cnt<rowtimes){for(int j=n-cnt;j>=cnt;j--){if(m-cnt<=m/2)break;retlist.add(matrix[m-cnt][j]);}}if(cnt<coltimes){for(int i=m-cnt-1;i>=cnt+1;i--){if(cnt+1>(n+1)/2)break;retlist.add(matrix[i][cnt]);}}}return retlist;}

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 ]]

public class Solution {    public static int[][] generateMatrix(int n){int times = (n + 1) / 2;int[][] matrix = new int[n][n];n--;int count = 1;for (int cnt = 0; cnt < times; cnt++){for (int j = cnt; j <= n - cnt; j++)if (cnt <= n)matrix[cnt][j] = count++;for (int i = cnt + 1; i <= n - cnt - 1; i++)if (n - cnt >= 0)matrix[i][n - cnt] = count++;for (int j = n - cnt; j >= cnt; j--){if (n - cnt <= n / 2)break;matrix[n - cnt][j] = count++;}for (int i = n - cnt - 1; i >= cnt + 1; i--){if (cnt + 1 > (n + 1) / 2)break;matrix[i][cnt] = count++;}}return matrix;}}