计蒜客---矩阵翻转(简单)

做这道题的过程中,了解到java的二维数组实际上就是多个一维数组的数组,所以类似下面这样的赋值是可以实现的.

    public static void upturn(int d[][],int a,int b) {
        int[] e = new int[b];
        for (int i = 0; i < a/2; i++) {
            e = d[i];
            d[i] = d[a-i-1];
            d[a-i-1] = e;
        }
    }

有了这个特性,那么矩阵翻转就变得很方便了,不过左右翻转,这个倒是没法实现

完整代码如下:

import java.util.Scanner;

public class Main{
    public static void main(String[] args){
        Scanner input = new Scanner(System.in);
        int a,b,c;
        a = input.nextInt();
        b = input.nextInt();
        c = input.nextInt();
        int[][] d = new int[a][b];
for (int i = 0; i < a; i++) {
    for (int j = 0; j < b; j++) {
        d[i][j] = input.nextInt();
    }
}
        if(c == 1){
            upturn(d, a, b);
        }else {
            leftturn(d, a, b);
        }
        for (int[] is : d) {
            for (int i : is) {
                System.out.print(i+" ");
            }
            System.out.println();
        }
    }
    public static void upturn(int d[][],int a,int b) {
        int[] e = new int[b];
        for (int i = 0; i < a/2; i++) {
            e = d[i];
            d[i] = d[a-i-1];
            d[a-i-1] = e;
        }
    }
    public static void leftturn(int d[][],int a,int b) {
        int temp;
        for (int i = 0; i < b/2; i++) {
            for (int j =0 ; j < a; j++) {
                temp = d[j][i];
                d[j][i] = d[j][b-i-1];
                d[j][b-i-1] = temp;
            }
        }
    }
}