[BASIC-25] 回形取数 - c++编程基础

基础练习回形取数时间限制：1.0s 内存限制：512.0MB 问题描述　　回形取数就是沿矩阵的边取数，若当前方向上无数可取或已经取过，则左转90度。一开始位于矩阵左上角，方向向下。输入格式　　输入第一行是两个不超过200的正整数m, n，表示矩阵的行和列。接下来m行每行n个整数，表示这个矩阵。输出格式　　输出只有一行，共mn个数，为输入矩阵回形取数得到的结果。数之间用一个空格分隔，行末不要有多余的空格。样例输入 3 3
1 2 3
4 5 6
7 8 9 样例输出 1 4 7 8 9 6 3 2 5 样例输入 3 2

1 2
3 4
5 6 样例输出 1 3 5 6 4 2

import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner scanner = new Scanner(System.in);

		while (scanner.hasNext()) {
			int m = scanner.nextInt();
			int n = scanner.nextInt();

			int[][] flag = new int[m][n];// 标记路径
			int[][] nums = new int[m][n];// 记录数据
			
			for (int i = 0; i < m; i++) {
				for (int j = 0; j < n; j++) {
					nums[i][j] = scanner.nextInt();
				}
			}

			int count = m * n;
			int i = 0, j = 0;

			while (count > 0) {
				while (i < m && flag[i][j] == 0) {
					System.out.print(nums[i][j]);
					System.out.print(--count == 0   "\r\n" : " ");
					flag[i][j] = 1;
					i++;
				}
				i--;
				j++;

				while (j < n && flag[i][j] == 0) {
					System.out.print(nums[i][j] + " ");
					System.out.print(--count == 0   "\r\n" : " ");
					flag[i][j] = 1;
					j++;
				}
				j--;
				i--;

				while (i > -1 && flag[i][j] == 0) {
					System.out.print(nums[i][j] + " ");
					System.out.print(--count == 0   "\r\n" : " ");
					flag[i][j] = 1;
					i--;
				}
				i++;
				j--;

				while (j > -1 && flag[i][j] == 0) {
					System.out.print(nums[i][j] + " ");
					System.out.print(--count == 0   "\r\n" : " ");
					flag[i][j] = 1;
					j--;
				}
				j++;
				i++;
			}
		}
	}
}