数组的初始化使用随机数赋值需要使用java.util.Random
//用0-100的数据初始化数组
public static void randomArry(double a[]) {
Random ran = new Random();
int n = a.length;
for(int i = 0;i <n ;i++) {
a[i] = ran.nextInt(100);
}
}
//初始化的重载
public static void randomArry(double a[][]) {
Random ran = new Random();
int n1 = a.length;
int n2 = a[0].length;
for(int i = 0;i<n1;i++) {
for(int j = 0;j < n2;j++) {
a[i][j] = ran.nextInt(100);
}
}
}寻找最大值
//找出数组的最大值
public static double findMax(double a[]) {
double max = a[0];
for(int i = 1;i < a.length;i++) {
if(a[i] >max) max = a[i];
}
return max;
}
//找最大值方法要创建一个对应的重载方法
public static double findMax(double a[][]) {
int n1 = a.length;
int n2 = a[0].length;
double max = a[0][0];
for(int i = 0; i < n1;i++) {
for(int j = 0; j < n2;j++) {
if(a[i][j] >max) max = a[i][j];
}
}
return max;
}计算数组元素的平均值
//计算数组元素的平均值
public static double figureAverage(double a[]) {
int n = a.length;
double sum = 0;
for(int i = 0; i < n;i++) {
sum += a[i];
}
return sum/n;
}
//当然我们算平均值也要算二维数组了看我们的重载方法如下
public static double figureAverage(double a[][]) {
int n1 = a.length;
int n2 = a[0].length;
double sum = 0;
for(int i = 0; i < n1;i++) {
for(int j = 0; j < n2;j++) {
sum += a[i][j];
}
}
return sum/(n1*n2);
}复制数组
//复制数组
public static double[] copyArry(double a[]) {
int n = a.length;
double b[] = new double[n];
for(int i = 0; i < n;i++) {
b[i] = a[i];
}
return b;
}
public static double[][] copyArry(double a[][]){
int n1 = a.length;
int n2 = a[0].length;
double[][] b = new double[n1][n2];
for(int i = 0; i<n1; i++) {
for(int j = 0; j < n2; j++) {
b[i][j] = a[i][j];
}
}
return b;
}颠倒数组元素的顺序
//颠倒数组元素的顺序
public static double[] switchElements(double a[]) {
int n = a.length;
for(int i = 0; i< n/2;i++) {
double temp = a[i];
a[i] = a[n - 1 -i];
a[n-1-i] = temp;
}
return a;
}
//颠倒数组顺序的重载方法
public static double[][] switchElements(double a[][]){
int n1 = a.length;
int n2 = a[0].length;
for(int i =0;i<n1/2;i++) {
for(int j = 0;j<n2;j++) {
if(i+j < (n1+n2)/2) {
double temp = a[i][j];
a[i][j] = a[n1-i-1][n2-j-1];
a[n1-i-1][n2-j-1] = temp;
}
}
}
return a;
}矩阵相乘
//矩阵相乘
public static double[] matrixMuiplication(double a[],double b[]) {
int n = a.length;
double[] c = new double[n];
for(int i = 0;i<n;i++) {
c[i] = a[i]*b[i];
}
return c;
}
//矩阵相乘的重载(方阵)
public static double[][] matrixMuiplication(double a[][],double b[][]) {
int n = a.length;
double c[][] = new double[n][n];
for(int i = 0;i < n;i++) {
for(int j = 0; j<n; j++) {
//计算行i和列j的点乘
for(int k = 0; k < n; k++) {
c[i][j] += a[i][k]*b[k][j];
}
}
}
return c;
}打印数组
//输出数组的方法
public static void printArry(double a[]) {
int n = a.length;
for(int i = 0; i<n ;i++) {
System.out.print(a[i]+"\t");
}
System.out.print("\n");
System.out.print("\n");
}
//数组输出的重载方法
public static void printArry(double a[][]) {
int n1 = a.length;
int n2 = a[0].length;
for(int i = 0; i<n1 ; i++) {
for(int j= 0;j<n2;j++) {
System.out.print(a[i][j]);
System.out.print("\t");
}
System.out.print("\n");
}
System.out.print("\n");
}完整代码带测试
1 import java.util.Random;
2
3 public class DianXingShuZhuChuLiDaiMa {
4
5
6 public static void main(String[] args) {
7 // 接下来对我们的数组方法逐一测试
8 double a[] = new double[8];
9 double b[] = new double[8];
10 double b1[][] = new double[8][8];
11 double b2[][] =new double[8][8];
12 printArry(a);
13 printArry(b);
14 printArry(b1);
15 printArry(b2);
16 //使用随机数来作为我们测试的量
17 randomArry(a);
18 randomArry(b);
19 randomArry(b1);
20 randomArry(b2);
21 //输出初始化的值
22 System.out.println("最初的a是");
23 printArry(a);
24 System.out.println("最初的b是");
25 printArry(b);
26 System.out.println("最初的b1是");
27 printArry(b1);
28 System.out.println("最初的b2是");
29 printArry(b2);
30
31 //找出a的最大值找出b2的最大值
32 System.out.println("a的最大值是:"+findMax(a));
33 System.out.println("b2的最大值是:"+findMax(b2));
34 //接下来计算数组元素的平均值
35 System.out.println("a的平均值是"+figureAverage(a));
36 System.out.println("b1的平均值是"+figureAverage(b1));
37 //复制数组
38 double d1[] = copyArry(a);
39 double d2[][] = copyArry(b1);
40 System.out.println("复制结果是");
41 printArry(d1);
42 printArry(d2);
43 //颠倒数组的元素顺序
44 System.out.println("颠倒数组的结果是a & b1");
45 printArry(switchElements(a));
46 printArry(switchElements(b1));
47
48 //测试矩阵相乘
49 System.out.println("ab相乘");
50 printArry(matrixMuiplication(a,b));
51 System.out.println("b1 b2相乘");
52 printArry(matrixMuiplication(b1,b2));
53
54
55 }
56
57 //用0-100的数据初始化数组
58 public static void randomArry(double a[]) {
59 Random ran = new Random();
60 int n = a.length;
61 for(int i = 0;i <n ;i++) {
62 a[i] = ran.nextInt(100);
63 }
64 }
65 //初始化的重载
66 public static void randomArry(double a[][]) {
67 Random ran = new Random();
68 int n1 = a.length;
69 int n2 = a[0].length;
70 for(int i = 0;i<n1;i++) {
71 for(int j = 0;j < n2;j++) {
72 a[i][j] = ran.nextInt(100);
73
74 }
75 }
76 }
77
78 //找出数组的最大值
79 public static double findMax(double a[]) {
80 double max = a[0];
81 for(int i = 1;i < a.length;i++) {
82 if(a[i] >max) max = a[i];
83 }
84 return max;
85 }
86 //找最大值方法要创建一个对应的重载方法
87 public static double findMax(double a[][]) {
88 int n1 = a.length;
89 int n2 = a[0].length;
90 double max = a[0][0];
91 for(int i = 0; i < n1;i++) {
92 for(int j = 0; j < n2;j++) {
93 if(a[i][j] >max) max = a[i][j];
94 }
95 }
96 return max;
97 }
98
99 //计算数组元素的平均值
100 public static double figureAverage(double a[]) {
101 int n = a.length;
102 double sum = 0;
103 for(int i = 0; i < n;i++) {
104 sum += a[i];
105 }
106 return sum/n;
107 }
108 //当然我们算平均值也要算二维数组了看我们的重载方法如下
109 public static double figureAverage(double a[][]) {
110 int n1 = a.length;
111 int n2 = a[0].length;
112 double sum = 0;
113 for(int i = 0; i < n1;i++) {
114 for(int j = 0; j < n2;j++) {
115 sum += a[i][j];
116 }
117 }
118 return sum/(n1*n2);
119 }
120
121 //复制数组
122 public static double[] copyArry(double a[]) {
123 int n = a.length;
124 double b[] = new double[n];
125 for(int i = 0; i < n;i++) {
126 b[i] = a[i];
127 }
128 return b;
129 }
130 public static double[][] copyArry(double a[][]){
131 int n1 = a.length;
132 int n2 = a[0].length;
133 double[][] b = new double[n1][n2];
134 for(int i = 0; i<n1; i++) {
135 for(int j = 0; j < n2; j++) {
136 b[i][j] = a[i][j];
137 }
138 }
139 return b;
140 }
141
142 //颠倒数组元素的顺序
143 public static double[] switchElements(double a[]) {
144 int n = a.length;
145 for(int i = 0; i< n/2;i++) {
146 double temp = a[i];
147 a[i] = a[n - 1 -i];
148 a[n-1-i] = temp;
149 }
150 return a;
151 }
152 //颠倒数组顺序的重载方法
153 public static double[][] switchElements(double a[][]){
154 int n1 = a.length;
155 int n2 = a[0].length;
156 for(int i =0;i<n1/2;i++) {
157 for(int j = 0;j<n2;j++) {
158 if(i+j < (n1+n2)/2) {
159 double temp = a[i][j];
160 a[i][j] = a[n1-i-1][n2-j-1];
161 a[n1-i-1][n2-j-1] = temp;
162 }
163 }
164 }
165 return a;
166 }
167
168 //矩阵相乘
169 public static double[] matrixMuiplication(double a[],double b[]) {
170 int n = a.length;
171 double[] c = new double[n];
172 for(int i = 0;i<n;i++) {
173 c[i] = a[i]*b[i];
174 }
175 return c;
176 }
177 //矩阵相乘的重载(方阵)
178 public static double[][] matrixMuiplication(double a[][],double b[][]) {
179 int n = a.length;
180 double c[][] = new double[n][n];
181 for(int i = 0;i < n;i++) {
182 for(int j = 0; j<n; j++) {
183 //计算行i和列j的点乘
184 for(int k = 0; k < n; k++) {
185 c[i][j] += a[i][k]*b[k][j];
186 }
187 }
188 }
189 return c;
190 }
191 //输出数组的方法
192 public static void printArry(double a[]) {
193 int n = a.length;
194 for(int i = 0; i<n ;i++) {
195 System.out.print(a[i]+"\t");
196 }
197 System.out.print("\n");
198 System.out.print("\n");
199 }
200 //数组输出的重载方法
201 public static void printArry(double a[][]) {
202 int n1 = a.length;
203 int n2 = a[0].length;
204 for(int i = 0; i<n1 ; i++) {
205 for(int j= 0;j<n2;j++) {
206 System.out.print(a[i][j]);
207 System.out.print("\t");
208 }
209 System.out.print("\n");
210 }
211 System.out.print("\n");
212 }
213 }输出结果
中国很多地方还需要提升，还有很多问题需要解决，如果十年后中国还没有发生改变，我觉得我就白做中国人了
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