leetcode 59 Spiral Matrix II 螺旋矩阵的打印

历时2个小时弄懂了螺旋矩阵的代码;

小结如下:

问题:

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 ]
]

 
 
这里以第一个元素为基准元素,不断寻找数组打印的边界index,同时每次更新基准元素,(row > starty*2 && column > startx*2)为边界条件,每次用endx = column - startx - 1;endy = row - starty - 1来更新下标


code:


 
 
public class Solution {
    public int[][] generateMatrix(int n){
		//means the max of index;
		int[][] ans = new int[n][n];
		if(n <= 0){
			System.out.println("print error!");
		}
		int row = n;
		int column = n;
		int startx = 0,starty = 0,endx,endy,i,j;
		int count = 1;
		
		while(row > starty*2 && column > startx*2) {
			
			endx = column - startx - 1;
			endy = row - starty - 1;
			
			//从左向右打印;
			for(j = startx; j <= endx; ++j) {
				ans[starty][j] = count; 
				++count;
			}
			
			//从上向下打印
			for(i = starty+1; i <= endy; ++i) {
				ans[i][endx] = count; 
				++count;
			}
			
			//从右向左打印
			for(j = endx-1; j >= startx; --j) {
				ans[endy][j] = count; 
				++count;
			}
			
			//从下向上打印
			for(i = endy-1; i >= starty+1; --i) {
				ans[i][startx] = count; 
				++count;
			}
			
			//基准元素发生变化;
			++startx;
			++starty;
		}
		return ans;
	}
}
AC
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