七十三.JAVA典型的数组处理

public class LianXi {
	public static void main(String[] args){
		//声明数组
		int a[];
		//创建数组
		a = new int[10];
		//初始化数组
		for(int i = 0; i < 10; i++){
			a[i] = 0;
		}
		//数组赋值
		for(int j = 0; j < 10; j++){
			a[j] = j;
		}
		
		int N = a.length;
		
		//找出数组最大的元素
		int max = a[0];
		for(int i = 0; i < N;i++){
			if(a[i]>max)
				max = a[i];
		}
		System.out.println("数组中的最大元素的值为：" + max);
		
		
		//计算数组元素的平均值
		double sum = 0.0;
		for(int i = 0; i < N; i++){
			sum += a[i];
		}
		double average = sum / N;
		System.out.println("数组元素的平均值为：" + average);
		
		
		//复制数组
		int [] b = new int [N];
		for(int i = 0 ; i < N; i++){
			b[i] = a[i];
		}
		System.out.print("数组b元素的值为：");
	      for(int i = 0; i < b.length;i++){
	          System.out.print(b[i] + " ");
	      }
	      System.out.println("\n");
	      
	      
		//颠倒数组元素的顺序
		for(int i = 0; i < N/2; i++){
			int temp = a[i];
			a[i] = a[N-1-i];
			a[N-1-i] = temp;
		}
		System.out.println("数组a颠倒后的值为：");
        for(int j = 0; j < N; j++){
			System.out.print(a[j] + " ");
		}
        System.out.println("\n");
        
        
		//矩阵相乘
		   int [][] a2 = {{1,2,3},{4,5,6},{7,8,9}};        // 3×3
	       int [][] b2 = {{1,2,3},{4,5,6},{7,8,9}};   // 3×3
	       int [][] c2 = new int [3][3];        // 3×3
	        for(int i = 0; i < 3; i++){
	            for(int j = 0; j < 3; j++){
	                for(int k = 0; k<3; k++){
	                    c2[i][j] = a2[i][k] * b2[k][j];
	                }
	            }
	        }
	        System.out.println("数组c2为：");
	        for(int i = 0; i<3; i++){
	        	for(int j = 0; j<3; j++){
	        		System.out.print(c2[i][j] + " ");
	        	}
	        	System.out.println("\n");
	        }
	}
}