经过一次正确的迭代后停止内循环

Question

代码中的问题出现在goldbach方法中。在最内层循环找到一对数字后，我想停止迭代内层的两个循环，但是我不知道如何退出这两个循环。换句话说，我只想在每个由最外面的for循环创建的整数中只找到一对，然后移动到下一个整数i。

下面是我的代码：

import java.util.Arrays; 
import java.awt.List; 
import java.util.ArrayList; 

// finding prime numbers using sieve of Eratosthenes and golbach's conjecture 
public class Test { 
    public static void main(String[] args) { 

     int[] num = new int[1000000]; 

     for (int i = 2; i <= num.length; i++) { 
      num[i - 1] = i; 
     } 
     Test.sieve(num); 


     Test.goldbach(num); 
    } 
    public static void sieve(int[] array) { 

     for (int i = 2; i < Math.sqrt(array.length); i++) { 
      if (array[i - 1] == 0) { 
       continue; 
      } 
      for (int j = 2 * i; j <= array.length; j += i) { 
       array[j - 1] = 0; 
      } 
     } 

     for (int i = 0; i < array.length; i++) { 
      if (array[i] != 0) { 
       //System.out.print(array[i] + " "); 
      } 
     } 
     //System.out.println(Arrays.toString(array)); 
    } 


    public static void goldbach(int[] array) { 
     for (int i = 2; i <= 1000000; i += 2) { //to go through every even integer 
      for (int j = 0; j <= i; j++) { 
       for (int k = 0; k <= i; k++) { 
        System.out.println("two prime numbers that add to " + i + " are " + array[j] + " and " + array[k]); 
        break; 
       } 
      } 
     } 
    } 
} 
} 

Answer 1

您可以设置j值在你的第二个循环。例如。

for (int i = 2; i <= 1000000; i += 2) { //to go through every even integer 
    for (int j = 0; j <= i; j++) { 
     for (int k = 0; k <= i; k++) { 
       System.out.println("two prime numbers that add to " + i + " are " + array[j] + " and " + array[k]); 
       j = i + 1; // This will end the outer loop as well. 
       break; 
      } 
     } 
    } 
} 

Answer 2

使用标签打破（或继续）比内另一个循环：

found: 
    for (int j = 0; j <= i; j++) { 
     for (int k = 0; k <= i; k++) { 
       System.out.println("two prime numbers that add to " + i + " are " + array[j] + " and " + array[k]); 
       break found; 
      } 
     } 
    } 

，或者使用一个额外的方法，并返回 - 更表示，如果新的方法有一个自己的清除功能（和更好名）

public static void goldbach(int[] array) { 
    for (int i = 2; i <= 1000000; i += 2) { //to go through every even integer 
     primeAdd(i); 
    } 
} 

private static void primeAdd(int i) { 
    for (int j = 0; j <= i; j++) { 
     for (int k = 0; k <= i; k++) { 
      System.out.println("two prime numbers that add to " + i + " are " + array[j] + " and " + array[k]); 
      return; 
     } 
    } 
} 

但是，作为已经被bureaquete评论说，没有必要为内循环，因为它总是被终止。