Example 1: (Read in an indeterminate integer from the keyboard, and judge the number of positive and negative numbers read, stop when you enter 0)
import java.util.Scanner;
class Test1{
public static void main(String[] args){
Scanner scan = new Scanner(System.in);
int positiveNumber = 0;//记录正数的个数
int negativeNumber = 0;//记录负数的个数
while(true){
//for(;;)
System.out.println("请输入一个整数：");
int number = scan.nextInt();
//判断number的正负情况
if(number > 0){
positiveNumber++;
}else if(number < 0){
negativeNumber++;
}else{
//一旦执行break，跳出循环
break;
}
}
System.out.println("正数个数为："+ positiveNumber);
System.out.println("负数个数为："+ negativeNumber);
}
}
to sum up:
1. The structure that does not limit the number of times in the loop condition part: for (;;) or while (true)
2. The way to end the loop:
①: return false in the loop condition part
②: execute break in the loop body;
Example 2: (Nine-Nine Multiplication Table)
class Test2 {
public static void main(String[] args){
for(int i = 1;i <= 9;i++){
for(int j = 1;j <= i;j++){
System.out.print(i + "*" + j + "=" + i*j + " ");
}
System.out.println();
}
}
}
Example 3: (traverse all prime numbers within 100)
method one:
class Test3 {
public static void main(String[] args) {
for(int i = 2;i <= 100;i++){
boolean isFlag = true;//标识i是否被j除尽过
for(int j = 2;j < i;j++){
if(i % j == 0){
isFlag = false;
}
}
if(isFlag == true){
System.out.println(i);
}
}
}
}
Way two:
class Test3 {
public static void main(String[] args) {
boolean isFlag = true;//标识i是否被j除尽过
for(int i = 2;i <= 100;i++){
for(int j = 2;j < i;j++){
if(i % j == 0){
isFlag = false;
}
}
if(isFlag == true){
System.out.println(i);
}
//重置isFlag
isFlag = true;
}
}
}
Method three: (optimization one) // unoptimized: 25886ms optimized: 5327ms
class Test3 {
public static void main(String[] args) {
boolean isFlag = true;//标识i是否被j除尽过
//获取当前时间距离1970-01-01 00:00:00 的毫秒数
long start = System.currentTimeMillis();
for(int i = 2;i <= 100000;i++){
for(int j = 2;j < i;j++){
if(i % j == 0){
isFlag = false;
break;//优化一，只对本身非质数的数有效
}
}
if(isFlag == true){
System.out.println(i);
}
//重置isFlag
isFlag = true;
}
//获取当前时间距离1970-01-01 00:00:00 的毫秒数
long end= System.currentTimeMillis();
System.out.println("所花费的时间为：" + (end - start));
//未优化：25886ms 优化：5327ms
}
}
Method four: (optimization two): optimization one: 5327ms optimization two: 1998ms
class Test3 {
public static void main(String[] args) {
boolean isFlag = true;//标识i是否被j除尽过
//获取当前时间距离1970-01-01 00:00:00 的毫秒数
long start = System.currentTimeMillis();
for(int i = 2;i <= 100000;i++){
//只需到根号i，对本身时质数的数有效
for(int j = 2;j <= Math.sqrt(i);j++){
if(i % j == 0){
isFlag = false;
break;//优化一，只对本身非质数的数有效
}
}
if(isFlag == true){
System.out.println(i);
}
//重置isFlag
isFlag = true;
}
//获取当前时间距离1970-01-01 00:00:00 的毫秒数
long end= System.currentTimeMillis();
System.out.println("所花费的时间为：" + (end - start));
//优化一：5327ms 优化二：1998ms
}
}
Method five: (optimization three) // optimization 2: 1998ms optimization 3: 21ms
class Test3
public static void main(String[] args) {
boolean isFlag = true;//标识i是否被j除尽过
int count = 0;//记录质数的个数
//获取当前时间距离1970-01-01 00:00:00 的毫秒数
long start = System.currentTimeMillis();
for(int i = 2;i <= 100000;i++){
//只需到根号i，对本身时质数的数有效
for(int j = 2;j <= Math.sqrt(i);j++){
if(i % j == 0){
isFlag = false;
break;//优化一，只对本身非质数的数有效
}
}
if(isFlag == true){
//System.out.println(i);
count++;
}
//重置isFlag
isFlag = true;
}
//获取当前时间距离1970-01-01 00:00:00 的毫秒数
long end= System.currentTimeMillis();
System.out.println("质数的个数为：" + count);
System.out.println("所花费的时间为：" + (end - start));
}
}
Method 6: Labeled break and continue
class Test3
public static void main(String[] args) {
int count = 0;//记录质数的个数
//获取当前时间距离1970-01-01 00:00:00 的毫秒数
long start = System.currentTimeMillis();
lable:for(int i = 2;i <= 100000;i++){
//只需到根号i，对本身时质数的数有效
for(int j = 2;j <= Math.sqrt(i);j++){
if(i % j == 0){
continue lable;
}
}
//能执行到此步骤的都是质数
count++;
}
//重置isFlag
isFlag = true;
}
//获取当前时间距离1970-01-01 00:00:00 的毫秒数
long end= System.currentTimeMillis();
System.out.println("质数的个数为：" + count);
System.out.println("所花费的时间为：" + (end - start));
}
}
