lambad表达式案例

1. 案例1：

    

    代码实现：

   package 黑马程序员;
   
   import java.util.Arrays;
   import java.util.Comparator;
   
   public class text {
   
   
       public static void main(String[] args) {
   
           girl gf1 = new girl(18, "aa", 1.67);
           girl gf2 = new girl(19, "bb", 1.7);
           girl gf3 = new girl(20, "cc", 1.8);
           girl[] girls={gf1,gf2,gf3};
           Arrays.sort(girls, ( o1,  o2) ->{
                   double temp=o1.getAge()-o2.getAge();
                   temp=temp==0?o1.getHeight()-o2.getHeight():temp;
                   temp=temp==0?o1.getName().compareTo(o2.getName()):temp;
                   if (temp>0){
                       return 1;
                   }else if (temp<0){
                       return -1;
   
                   } else {
                       return 0;
                   }
               }
           );
           System.out.println(Arrays.toString(girls));
       }
   
   
       }
       class girl{
       private int age;
       private String name;
       private double height;
   
           public girl(int age,String name, double  height) {
               this.age = age;
               this.name = name;
               this.height = height;
           }
   
           public girl() {
           }
   
           public int getAge() {
               return age;
           }
   
           public void setAge(int age) {
               this.age = age;
           }
   
           public String getName() {
               return name;
           }
   
           public void setName(String name) {
               this.name = name;
           }
   
           public double getHeight() {
               return height;
           }
   
           public void setHeight(int height) {
               this.height = height;
           }
       }
   
   

    

2. 案例2：

    

    

3.  

   案例2代码实现：

   public class text2 {
       public static void main(String[] args) {
           System.out.println(getway(20));
       }
       public  static int getway(int number){
           if (number==1){
               return 1;
   
           }
           if (number==2){
               return 2;
   
           } else {
               return getway(number-1)+getway(number-2);
           }
       }
   }

4. 斐波那契数列（不死神兔问题）：特点是第3个数据是前两个数据之和

    

    

    

5.  

   代码实现： 

   public class text2 {
       public static void main(String[] args) {
           //方法1:迭代
           int[] number=new int[12];
           number[0]=1;
           number[1]=1;
           for (int i = 2; i <number.length ; i++) {
               number[i]=number[i-1]+number[i-2];
               
   
   
           }
           System.out.println(number[11]);
         
       }
      
   }

6. 代码实现2：

   public class text2 {
       public static void main(String[] args) {
          //方法2：递归
           System.out.println(getWay(12));
   
       }
       public static int getWay(int number){
           if (number==1||number==2){
               return 1;
   
           }else {
               return getWay(number-1)+getWay(number-2);
   
           }    }
   }