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
| 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 | package base25; import java.util.Scanner; public class Main {     public static void main(String[] args) {         Scanner in = new Scanner(System.in);         int n = in.nextInt();         int m = in.nextInt();         int[][] a = new int[200][200];         for (int k = 0; k < n; k++) {             for (int k1 = 0; k1 < m; k1++) {                 a[k][k1] = in.nextInt();             }         }         int all = n * m;         int i = 0, j = 0;         while (all > 0) {             while (i < m && a[i][j] != -1) {//down                 System.out.print(a[i][j] + " ");                 a[i][j] = -1;                 i++;                 all--;             }             i--;             j++;             while (j < n && a[i][j] != -1) {//right                 System.out.print(a[i][j] + " ");                 a[i][j] = -1;                 j++;                 all--;             }             j--;             i--;             while (i >= 0 && a[i][j] != -1) {//up                 System.out.print(a[i][j] + " ");                 a[i][j] = -1;                 i--;                 all--;             }             i++;             j--;             while (j >= 0 && a[i][j] != -1) {//left                 System.out.print(a[i][j] + " ");                 a[i][j] = -1;                 j--;                 all--;             }             j++;             i++;         }     } } | 
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