leetcode--Spiral Matrix II

Given an integer n, generate a square matrix filled with elements from 1 ton² in spiral order.

For example,
Given n = 3,

You should return the following matrix:

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

题意：给定n,得到n*n的环形数组

分类：数组

解法1：和leetcode--Spiral Matrix一样，只是访问过程变成赋值

public class Solution {
    public int[][] generateMatrix(int n) {
        int[][] res = new int[n][n];
		int count = 1;
        int rows = n;    
        if(rows==0) return res;  
        int cols = n;    
        boolean vis[][] = new boolean[rows][cols];    
        int x = 0;                  
        while(x*2<cols && x*2<rows){    
            int endX = rows-x-1;    
            int endY = cols-x-1;    
            for(int i=x;i<=endY;i++){    
                if(!vis[x][i]){    
                    res[x][i]=count++;    
                    vis[x][i] = true;    
                }    
            }    
            for(int i=x+1;i<=endX;i++){    
                if(!vis[i][endY]){    
                	res[i][endY]=count++;    
                    vis[i][endY] = true;    
                }    
            }    
            for(int i=endY-1;i>=x;i--){    
                if(!vis[endX][i]){    
                	res[endX][i]=count++;    
                    vis[endX][i] = true;    
                }    
            }    
            for(int i=endX-1;i>x;i--){    
                if(!vis[i][x]){    
                	res[i][x]=count++;
                    vis[i][x] = true;    
                }                   
            }               
            x = x + 1;              
        }    
        return res;    
    }
}

解法2：优化了遍历过程

public class Solution {
    public int[][] generateMatrix(int n) {
        int[][] res = new int[n][n];
        if(n==0) return res;  
        int x = 0,y = 0;  
        int count = 0,limit = n*n;  
        int right = n-1,bottom = n-1,left = 0,top = 0;  
        while(count<limit){  
            while(count<limit&&y<=right){res[x][y++]=++count;}  
            y--;x++;top++;  
            while (count<limit&&x<=bottom) {res[x++][y]=++count;}  
            x--;y--;right--;  
            while(count<limit&&y>=left) {res[x][y--]=++count;}  
            x--;y++;bottom--;  
            while(count<limit&&x>=top) {res[x--][y]=++count;}  
            x++;y++;left++;  
        }  
        return res;  
    }
}