java : 递归实例

当原方法的实现和子方法相同,并且子方法进一步简化了原方法(执行步骤减少了,判断的次数减少了等),可以考虑使用递归处理问题

1.使用递归实现斐波那契数列

View Code
 1 //使用递归实现斐波那契数列
 2 import java.util.* ;
 3 public class TestFibonacci
 4 {
 5    public static void main(String[] args)
 6    {
 7        Scanner input = new Scanner(System.in) ;
 8        int count = 0 ;
 9        int num ;
10        System.out.print("请输入斐波那契数列的个数: ") ;
11        num = input.nextInt() ;
12        for(int i = 1 ; i<=num ;i++)
13        {
14            System.out.print(fibonacci(i)+" ") ;
15            count++ ;
16            if(count%10==0)
17               System.out.println() ;
18         }
19     }
20    public static int fibonacci(int num)
21    {
22        if(num==1||num==2)
23            return 1 ;
24        else
25            return (fibonacci(num-2)+fibonacci(num-1)) ;
26    }
27 }

2.使用递归实现阶乘

View Code
 1 //使用递归实现阶乘
 2 import java.util.* ;
 3 public class TestFactorial
 4 {
 5    public static void main(String[] args)
 6    {
 7        Scanner input = new Scanner(System.in) ;
 8        int n ;
 9        System.out.print("请输入阶乘n!中的n :") ;
10        n = input.nextInt() ;
11        int result = factorial(n) ;
12        System.out.println("The result of "+n+"! is "+result) ;
13     }
14     public static int factorial(int n)
15     {
16         if(n==1)
17            return 1 ;
18         else
19            return n*factorial(n-1) ;
20     }
21 }

3.使用递归打印N次消息

View Code
 1 //使用递归打印N次消息
 2 public class TestRecursionOutString
 3 {
 4     public static void main(String[] args)
 5     {
 6         int time = 10 ;
 7         nPrint("I love you.",time) ;
 8     }
 9     public static void nPrint(String message,int time)
10     {
11         if(time>=1)
12            System.out.println(message) ;
13         if(time-1>0)
14            nPrint(message,time-1) ;
15     }
16 }

4.使用递归实现回文判断

View Code
 1 //使用递归实现回文判断
 2 import java.util.* ;
 3 public class TestIsPalindrome
 4 {
 5    public static void main(String[] args)
 6    {
 7        while(true)
 8        {
 9           Scanner input = new Scanner(System.in) ;
10           System.out.print("请输入一个字符串(判断是否为回文符,输入'end'时,推出程序): ") ;
11           String palindrome = input.next() ;
12           if(palindrome.equals("end")==true)
13                break ;
14           else
15              System.out.println(palindrome+" is a palindrome ? "+isPalindrome(palindrome)) ;
16         }
17     }
18    public static boolean isPalindrome(String palindrome)
19    {
20        if(palindrome.length()==1)
21          return true ;
22        else
23           if(palindrome.length()==2)
24           {
25               if(palindrome.charAt(0)==palindrome.charAt(palindrome.length()-1))
26                  return true ;
27               else
28                  return false ;
29           }
30           else
31              if(palindrome.charAt(0)!=palindrome.charAt(palindrome.length()-1))
32                 return false ;
33              else
34                 return isPalindrome(palindrome.substring(1,palindrome.length()-1)) ;
35     }
36 }

递归辅助方法

5.通过递归辅助方法重写回文符的判断

View Code
 1 //通过递归辅助方法重写回文符的判断
 2 import java.util.* ;
 3 public class TestIsPalindrome2
 4 {
 5      public static void main(String[] args)
 6    {
 7        while(true)
 8        {
 9           Scanner input = new Scanner(System.in) ;
10           System.out.print("请输入一个字符串(判断是否为回文符,输入'end'时,推出程序): ") ;
11           String palindrome = input.next() ;
12           if(palindrome.equals("end")==true)
13                break ;
14           else
15              System.out.println(palindrome+" is a palindrome ? "+isPalindrome2(palindrome)) ;
16         }
17     }
18     public static boolean isPalindrome2(String s)
19     {
20         return isPalindrome2(s,0,s.length()-1) ;
21     }
22     public static boolean isPalindrome2(String s,int low,int high)
23     {
24         if(high<=low)
25            return true ;
26         else
27            if(s.charAt(low)!=s.charAt(high))
28               return false ;
29            else
30               return isPalindrome2(s,low+1,high-1) ;
31     }
32 }

6.通过递归实现选择排序

View Code
 1 //通过递归实现选择排序
 2 public class TestSort
 3 {
 4     public static void main(String[] args)
 5     {
 6         int[] nums = {2,6,7,3,9,5} ;
 7         display(nums) ;
 8         sort(nums) ;
 9         display(nums) ;
10     }
11     public static void sort(int[] nums)
12     {
13         sort(nums,nums.length-1) ;
14     }
15     public static void sort(int[] nums,int high)
16     {
17         if(high>=1)
18         {
19             int maxIndex = 0 ;
20             int max = nums[0] ;
21             for(int i=1 ; i<=high ; i++)
22             {
23                 if(max < nums[i])
24                 {
25                     maxIndex = i ;
26                     max = nums[i] ;
27                 }
28             }
29             nums[maxIndex] = nums[high] ;
30             nums[high] = max ;
31             sort(nums,high-1) ;
32         }
33     }
34     public static void display(int[] nums)
35     {
36         for(int i=0 ;i<nums.length ; i++)
37            System.out.print(nums[i]+" ") ;
38         System.out.println() ;
39     }
40 }

7.使用递归实现二分查找

View Code
 1 //使用递归实现二分查找
 2 public class TestBinarySearch
 3 {
 4    public static void main(String[] args)
 5    {
 6        int[] nums = {2,4,6,8,10,12,14,16} ;
 7        display(nums) ;
 8        int key = 8 ;
 9        int searchNum = binarySearch(nums,key) ;
10        System.out.println(key+" 所在的下标为: "+searchNum);
11     }
12    public static int binarySearch(int[] nums,int key)
13    {
14        int low = 0 ;
15        int high = nums.length-1 ;
16        return binarySearch(nums,key,low,high) ;
17     }
18    public static int binarySearch(int[] nums,int key,int low,int high)
19    {   
20        if(low>high)
21           return -low-1 ;
22        int mid = (low+high)/2 ;
23        if(nums[mid]==key)
24           return mid ;
25        else 
26           if(nums[mid]>key)
27              return binarySearch(nums,key,low,mid-1) ;
28           else
29              return binarySearch(nums,key,mid+1,high) ;
30    }
31    public static void display(int[] nums)
32     {
33         for(int i=0 ;i<nums.length ; i++)
34            System.out.print(nums[i]+" ") ;
35         System.out.println() ;
36     }
37 }

8.用递归解决汉诺塔问题

View Code
 1 //用递归解决汉诺塔问题
 2 public class TowerOfHanoi
 3 {
 4    public static void main(String[] args)
 5    {
 6       int n = 3 ;
 7       System.out.println("The result is ") ;
 8       moveDisks(n,'A','B','C') ;//把n个盘从借助C盘,从A盘搬到B盘
 9     }
10    public static void moveDisks(int n,char from,char to,char help)
11    {
12        if(n==1)
13           System.out.println("Move disk "+n+" from "+from+" to "+to) ;
14        else
15        {
16            moveDisks(n-1,from,help,to) ;//将n-1个盘借助B盘,从A盘搬到C盘
17            System.out.println("Move disk "+n+" from "+from+" to "+to) ;//将N盘从A盘搬到B盘
18            moveDisks(n-1,help,to,from) ;//将n-1个盘借助A盘,从C盘搬到B盘
19        }
20     }
21 }

9.用递归求得两个整数的最大公约数

View Code
 1 //求连个整数的最大公约数(使用递归)
 2 //使用辗转相除法:
 3 import java.io.* ; 
 4 public class ComDivisor
 5 {
 6     public static void main(String[] agrs) throws IOException
 7     {
 8         System.out.print("请输入第一个整数: ") ;
 9         int a = getInt() ;
10         System.out.print("请输入第二个整数: ") ;
11         int b = getInt() ;
12         int result = getComDivisor(a,b) ;
13         System.out.println(a+","+b+"最大的公约数是:"+result) ;
14     }
15 //------------------------------------------
16     public static int getComDivisor(int a,int b)//返回最大公约数
17     {
18         if(a<b)
19            return doComDivisor(a,b) ;
20         else
21            return doComDivisor(b,a) ;
22     }
23     public static int doComDivisor(int a,int b)//小数在前,大数在后
24     {
25         if(b%a==0)
26            return a ;
27         else
28         {
29             int c = b%a ;
30             return doComDivisor(c,a) ;
31         }
32     }
33 //------------------------------------------
34     public static String getString() throws IOException
35    { 
36       InputStreamReader is = new InputStreamReader(System.in) ;
37       BufferedReader br = new BufferedReader(is) ;
38       String s = br.readLine() ;
39       return s ;
40     }
41 //------------------------------------------
42     public static int getInt() throws IOException
43    { 
44       String s = getString() ;
45       return Integer.parseInt(s) ;
46     }
47 }

 

 

 

 

 

 

 

原文地址:https://www.cnblogs.com/KeenLeung/p/2703503.html