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;
	}
}



评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值