13518219792
import java.util.Scanner;
创新互联基于分布式IDC数据中心构建的平台为众多户提供绵阳电信机房机柜租用 四川大带宽租用 成都机柜租用 成都服务器租用。
class Example
{
/*递归方法*/
public long square(int n)
{
if (0 == n)//递归的出口
{
return 0;
}
else
{
return square(n-1) + 2*n - 1;
}
}
/*输入数据n的方法*/
public int scan()
{
int number = 0;
Scanner scan = null;
while(true)
{
System.out.print("请输入 n 的值：");
try{
scan = new Scanner(System.in);
number = scan.nextInt();
break;
}catch(Exception e)
{
System.out.println("你输入的数据n 错误!");
System.out.println("请重新输入...");
}
}
return number;
}
public boolean isContinue ()
{
String src = null;
Scanner scan = null;
while(true){
System.out.print("是否还需要继续计算？y(yes)/n(no) : ");
scan = new Scanner(System.in);
src = scan.nextLine();
if(src.equalsIgnoreCase("y") || src.equalsIgnoreCase("yes"))
{
return true;
}
else if(src.equalsIgnoreCase("n") || src.equalsIgnoreCase("no"))
{
return false;
}
}
}
public static void main(String[] args)
{
int number = 0;
Example exam = null;
do{
exam = new Example();
number = exam.scan();
long result = exam.square(number);
System.out.println("square(" + number + ") = " + result);
}while(exam.isContinue());
}
}
这个很好写的，代码如下：
private ListDept recursionDept(ListDept ld){
for(int i=0; ild.size(); i++) {
Dept d = ld.get(i)
loop(d);
}
}
private void loop(Dept d) {
ListDept children=service.getChildDept(d.id);
if (children.size() 0) {
d.setChildren(children); // 这里假设子列表属性的名字就是children
for(int j=0; jchidren.size(); j++){
loop(children.get(j);
}
}
}
这个题目对初学者来说比较难的一点是，得想明白要自己建一个递归方法（loop）
阶乘：
要求：给定一个数值，计算出它的阶乘值，例如5的阶乘为5*4*3*2*1
实现：
[html] view plaincopy
span style="font-size:12px;" // 利用递归实现一个数的阶乘值 private static BigDecimal getNum(BigDecimal inNum) { if (inNum.compareTo(BigDecimal.ONE) == 0) { return inNum; } return inNum.multiply(getNum(inNum.subtract(BigDecimal.ONE))); }/span
（2）Fibonacci数列：1，1，2，3，5，8，13……
要求：找出数列中指定index位置的数值
实现：
[html] view plaincopy
span style="font-size:12px;" // 利用递归实现了Fibonacci数列 private static int fab(int index) { if (index == 1 || index == 2) { return 1; } else { return fab(index - 1) + fab(index - 2); } }/span
（3）汉诺塔
要求：汉诺塔挪动
实现：
[html] view plaincopy
span style="font-size:12px;" span style="white-space:pre;" /spanprivate static final String DISK_B = "diskB"; span style="white-space:pre;" /spanprivate static final String DISK_C = "diskC"; span style="white-space:pre;" /spanprivate static final String DISK_A = "diskA"; span style="white-space:pre;" /spanstatic String from=DISK_A; span style="white-space:pre;" /span static String to=DISK_C; span style="white-space:pre;" /span static String mid=DISK_B; span style="white-space:pre;" /span public static void main(String[] args) { span style="white-space:pre;" /span String input=JOptionPane.showInputDialog("please input the number of the disks you want me move."); span style="white-space:pre;" /span int num=Integer.parseInt(input); span style="white-space:pre;" /span move(num,from,mid,to); span style="white-space:pre;" /span }/span
[html] view plaincopy
span style="font-size:12px;" // 利用递归实现汉诺塔 private static void move(int num, String from2, String mid2, String to2) { if (num == 1) { System.out.println("move disk 1 from " + from2 + " to " + to2); } else { move(num - 1, from2, to2, mid2); System.out.println("move disk " + num + " from " + from2 + " to " + to2); move(num - 1, mid2, from2, to2); } }/span
（4）排列组合
要求：将输入的一个字符串中的所有元素进行排序并输出，例如：你给出的参数是"abc"，
则程序会输出
abc
acb
bac
bca
cab
cba
实现：
[html] view plaincopy
span style="font-size:12px;"span style="white-space:pre;" /spanpublic static void permute(String str) { span style="white-space:pre;" /span char[] strArray = str.toCharArray(); span style="white-space:pre;" /span permute(strArray, 0, strArray.length - 1); span style="white-space:pre;" /span}/span
[html] view plaincopy
span style="font-size:12px;" // 利用递归实现，将输入的一个字符串中的所有元素进行排序并输出 public static void permute(char[] list, int low, int high) { int i; if (low == high) { String cout = ""; for (i = 0; i = high; i++) { cout += list[i]; } System.out.println(cout); } else { for (i = low; i = high; i++) { char temp = list[low]; list[low] = list[i]; list[i] = temp; permute(list, low + 1, high); temp = list[low];
代码如下：
import java.util.ArrayList;
import java.util.List;
class Org {
private String id;
private String name;
private String pid;
public Org(String id, String name, String pid) {
this.id = id;
this.name = name;
this.pid = pid;
}
public String getId() {
return id;
}
public void setId(String id) {
this.id = id;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public String getPid() {
return pid;
}
public void setPid(String pid) {
this.pid = pid;
}
@Override
public String toString() {
return "Org [id=" + id + ", name=" + name + ", pid=" + pid + "]";
}
}
public class App {
static void find(ListOrg list, String pid) {
list.stream().filter(p - p.getPid().equals(pid))
.forEach(org - {
System.out.println(org);
find(list, org.getId());
});
}
public static void main(String[] args) {
ListOrg list = new ArrayList();
list.add(new Org("111", "公司", "0"));
list.add(new Org("222", "部门", "111"));
list.add(new Org("333", "小组", "222"));
list.add(new Org("444", "员工1", "333"));
list.add(new Org("555", "员工2", "333"));
find(list, "0");
System.out.println("------------------------------------");
find(list, "111");
}
}
运行结果：
下面递归写了一段递归累加到100，每加20个就换行输出。
package zhidao;
public class Digui {
public static int add(int num){
int sum = 0;
StringBuffer sb = new StringBuffer();
if (num = 0) {
return 0;
}else{
if (num == 1) {
sum = sum+1;
}else {
sum = add(num-1)+num;
}
if (num % 20 == 0) {
System.out.println("[index = "+num+" sum = "+sum+"]");
}else {
System.out.print("[index = "+num+" sum = "+sum+"],");
}
}
return sum;
}
public static void main(String[] args) {
add(100);
}
}
