数组定义

题目1：

public class Hw1{public static void main(String[] args){char[] c1 = new char[26];c1[0] = 'A';for (int i = 1; i <= 25; i++){c1[i] = (char)(c1[0] + i);}System.out.println(c1);
//自己写的烂方法：char[] c2 = new char[26];for (int i = 0; i < c2.length; i++){c2[i] = 'A' + i;}System.out.println(c2);
// 老师给的牛犇的方法}
}

题目2：

public class Hw1{public static void main(String[] args){int[] a = {4, -1, 9, 10, 23};int max = a[0];int indx = 0;for (int i = 0; i < a.length; i++){if (max <= a[i]){max = a[i];indx = i;}}System.out.println("最大值是"+max+" 下标是"+indx);}
}

数组赋值

数组的拷贝

将int[] arr1 = {1, 2, 3} 拷贝到arr2数组，要求数据空间是独立的

1. 创立新的数组，开辟新空间

2. 遍历arr1，把每个元素拷贝到arr2的对应位置

数组翻转

思路1：注意吧len = arr.length

public class Hw1{public static void main(String[] args){int[] arr = {11, 22, 33, 44, 55, 66};//把arr[1]和arr[5]交换，//  arr[2]和arr[4]交换...int len = arr.length;for(int i = 0; i < len / 2; i++){int temp = arr[i];arr[i] = arr[len - i - 1];arr[len - i - 1] = temp;}for(int i = 0; i < len; i++){System.out.println(arr[i]);}}
}

思路2：逆序赋值法

1. 先开辟一个新的数组

2. 逆序遍历arr1, 顺序遍历新数组，把arr1的那个数据反着传给新数组

3. 记得要把arr2的地址给到arr1，不然arr1没改变，然后会被java垃圾回收机制收走

4. 注意：主要是那个arr2的下标用j，仔细看一下

public class Hw1{public static void main(String[] args){int[] arr1 = {11, 22, 33, 44, 55, 66};int[] arr2 = new int[arr1.length];for(int i = arr1.length - 1, j = 0; i >= 0; i--, j++){arr2[j] = arr1[i];}arr1 = arr2;for (int i = 0; i < arr2.length; i++){System.out.print(arr1[i]+"\t");}}
}

数组扩容

要求：给arr添加内容，先问用户要不要添加，y/n

思路：

1. 开辟新数组，长度是原数组+1，把原数组依次赋值

2. 在屁股上加一个

3. 改指针

4. 用do-while + break控制用户要不要添加

import java.util.Scanner;
public class Lianxi{public static void main(String[] args){int arr[] = {1, 2, 3};do{Scanner myScanner = new Scanner(System.in);System.out.println("请问您是否要为数组添加一个数据？");char key = myScanner.next().charAt(0);if (key == 'n'){break;}int[] arr2 = new int[arr.length + 1];System.out.println("请输入您要添加的数据？");int num = myScanner.nextInt();for (int i = 0; i < arr.length; i++){arr2[i] = arr[i];}arr2[arr2.length - 1] = num;arr = arr2;for (int i = 0; i < arr2.length; i++){System.out.print(arr2[i] + "\t");}} while(true);}
}

数组缩减

排序

冒泡排序：外层控制我们一共要比较多少轮，内层控制从第0个数开始，要比多少次

import java.util.Scanner;
public class Lianxi{public static void main(String[] args){int arr[] = {2, 10, 7, 5, 8, 6, 9};for(int i = 0; i < arr.length - 1; i++){for (int j = 0; j < arr.length - 1 - i; j++){if (arr[j] >= arr[j+1]){int temp = arr[j];arr[j] = arr[j + 1];arr[j + 1] = temp;}}}for(int i = 0; i < arr.length; i++){System.out.print(arr[i] + "\t");}}
}

查找

注意：有个小技巧，初始化index=-1来表示没有找到匹配的

import java.util.Scanner;
public class Lianxi{public static void main(String[] args){String[] names = {"hy", "lwj", "dxy", "tl", "zsq"};Scanner myScanner = new Scanner(System.in);System.out.println("请输入您要查找的字符串：");String name = myScanner.next();//技巧：index = -1 int index = -1;for (int i = 0; i < names.length; i++){if(name.equals(names[i])){index = i;System.out.println(name + "是第" + i +"个字符串");}}if(index == -1){System.out.println("没找到" + name);}}
}

二维数组

定义：原来每个一维数组的每个元素是一维数组，就构成二维数组

第一个表示二维数组里面有几个一维数组，第二个表示每个一维数组里面有几个元素

二维数组在内存里面的存在形式：

ok，现在打印出图片中的数组：

public class Lianxi{public static void main(String[] args){int[][] arr = new int[3][];//新建一个二维数组，有三个一维数组构成，但是每个一维数组没有开空间，指向的地址都是nullfor(int i = 0; i < arr.length; i++){arr[i] = new int[i + 1];//给每个一维数组单独开辟空间for(int j = 0; j < arr[i].length; j++){arr[i][j] = i + 1;System.out.print(arr[i][j] + " ");}System.out.println("");}}
}

静态初始化：

二维数组的应用

杨辉三角形

/*打印杨辉三角形
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
*/
//主要是三个规律把握住：
/*(1) 第i+1行有i个元素(2) 每行首末数字为1(3) yanghui[i][j] = yanghui[i - 1][j] + yanghui[i - 1][j - 1];
*/
public class Lianxi{public static void main(String[] args){int[][] yanghui = new int[10][];for(int i = 0; i < yanghui.length; i++){yanghui[i] = new int[i + 1];yanghui[i][0] = 1;yanghui[i][yanghui[i].length - 1] = 1;for(int j = 1; j < yanghui[i].length - 1; j++){yanghui[i][j] = yanghui[i - 1][j] + yanghui[i - 1][j - 1];}}for(int i = 0; i < yanghui.length; i++){for(int j = 0; j < yanghui[i].length; j++){System.out.print(yanghui[i][j]+" ");}System.out.println("");}}
}

本章练习题

1

我的思路：

先定位，再插入，

定位：采用index来定，这个和老师想得一样，但是遍历结束a， index=-1 之后，更为简洁的是，让index = a.length 表示插入的位置在最后一个。

插入：我的想法只用了一个下标，所以代码看着挺复杂的，没有老师的简洁。

import java.util.Scanner;
public class Lianxi{public static void main(String[] args){int[] a = {10, 12, 45, 99};int[] b = new int[a.length + 1];Scanner myScanner = new Scanner(System.in);System.out.println("请输入您要添加的数字");int num = myScanner.nextInt();int index = -1;for(int i = 0; i < b.length; i++){if(a[i] >= num){index = i;break;}}for(int i = 0; i < b.length - 1; i++){if (index == -1){b[i] = a[i];b[b.length - 1] = num;}else if (i < index){b[i] = a[i];}else if(i == index){b[i] = num;b[i + 1] = a[i];}else if( i > index){b[i + 1] = a[i];}}for(int i = 0; i < b.length; i++){System.out.print(b[i] + " ");}}
}

下面附上老师的思路：

用i表示新数组下标，j表示原来的数组下标，赋值的时候跳过index这个地方

import java.util.Scanner;
public class Lianxi{public static void main(String[] args){int[] a = {10, 12, 45, 99};int[] b = new int[a.length + 1];Scanner myScanner = new Scanner(System.in);System.out.println("请输入您要添加的数字");int num = myScanner.nextInt();int index = -1;for(int i = 0; i < b.length; i++){if(a[i] >= num){index = i;break;}}if (index == -1){index = a.length;}for(int i = 0, j = 0; i < b.length; i++){if(index != i){b[i] = a[j];j++;}else{b[i] = num;}}for(int i = 0; i < b.length; i++){System.out.print(b[i] + " ");}}
}

import java.util.Scanner;
public class Lianxi{public static void main(String[] args){int[] a = new int[10];for (int i = 0; i < a.length; i++){a[i] = (int)(Math.random()*100) + 1;}double sum = 0;int max = a[0];int maxIndex = -1;for(int i = a.length - 1; i >= 0; i --){System.out.print(a[i] + "\t");sum += a[i];if(max <= a[i]){max = a[i];maxIndex = i;}if(a[i] == 8){System.out.println(a[i] + "= 8");}}System.out.println("最大值为" + max + "，下标为" + maxIndex);System.out.println("平均值为" + sum/a.length);}
}

冒泡排序：

import java.util.Scanner;
public class Lianxi{public static void main(String[] args){int[] a ={3, 6, 1, 5, 2, 10, 9};int temp = 0;for (int i = 0; i < a.length - 1 ; i++){for (int j = 0; j < a.length - i - 1; j++){if(a[j] >= a[j + 1]){temp = a[j];a[j] = a[j + 1];a[j + 1] = temp;}}}for(int i = 0; i < a.length; i++){System.out.print(a[i] + " ");}}
}