【Leetcode】Spiral Matrix 一 和 二 in JAVA

首先是1：

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].

这里有一个问题，由于m和n很有可能不一样，所以不一样while转一圈才能跳出循环，所以每一个for循环后要判断是否要跳出break。

package testAndfun;

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

public class spriralOrder {
	List<Integer> list = new ArrayList<Integer>();
	public static void main(String[] args){
		spriralOrder so = new spriralOrder();
		int[][] m = {{1,2,3},{4,5,6},{7,8,9}};
		//int[][] m = {{2,3}};
		System.out.println(so.spiralOrders(m).toString());
		
	}
	public List<Integer> spiralOrders(int[][] matrix) {
		if(matrix.length==0 || matrix[0].length==0)	return list;
        int moveStep = 0;
        int m = matrix[0].length;
        int n = matrix.length;
        int x=0,y=-1;
        while(moveStep!=matrix.length*matrix[0].length){
        	for(int i=0;i<m;i++){
        		list.add(matrix[x][++y]);
        		moveStep++;
        	}
        	if(moveStep==matrix.length*matrix[0].length)	break;
        	for(int i=0;i<n-1;i++){
        		list.add(matrix[++x][y]);
        		moveStep++;
        	}
        	if(moveStep==matrix.length*matrix[0].length)	break;
        	for(int i=0;i<m-1;i++){
        		list.add(matrix[x][--y]);
        		moveStep++;
        	}
        	if(moveStep==matrix.length*matrix[0].length)	break;
        	for(int i=0;i<n-2;i++){
        		list.add(matrix[--x][y]);
        		moveStep++;
        	}
        	n = n-2;
        	m = m-2;
        }
        return list;
    }
}

然后是2：

Given an integer n, generate a square matrix filled with elements from 1 to n2 in spiral order.

For example,
Given n = 3,

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

package testAndfun;

import java.util.Arrays;

public class SpiralMatrix {
	public static void main(String[] args){
		SpiralMatrix sm = new SpiralMatrix();
		System.out.println(Arrays.deepToString(sm.generateMatrix(3)));
	}
public int[][] generateMatrix(int n) {
		int[][] out = new int[n][n];
        if(n<=0) return out;
        int x=0;
        int y=-1;
        int moveStep = 0;
        while(moveStep != n*n){
        	for(int i=0;i<n;i++){
        		out[x][++y] = ++moveStep;
        	}
        	for(int i=0;i<n-1;i++){
        		out[++x][y] = ++moveStep;
        	}
        	for(int i=0;i<n-1;i++){
        		out[x][--y] = ++moveStep;
        	}
        	for(int i=0;i<n-2;i++){
        		out[--x][y] = ++moveStep;
        	}

        	n = n-2;
        }
        return out;
}
}