嵌套循环执行

Question

这是一个非常简单的问题，但出于某种原因我感到困惑而且很烦人。

def test():
for i from Lo1 to Hi1:
    for j from Lo2 to Hi2:
    body()

body() 将执行多少次： Lo1=1, Hi1=n, Lo2=i-2, Hi2=i+2

答案是5n次，但我不知道他们是怎么得到的！

Answer 1

Lo2=i-2, Hi2=i+2 内环

So (i-2)to(i+2)=5 Alternations EG: i-2 , i-1 , i , i+1 , i+2

Like (-2) to (+2)=-2,-1,0,1,2

Lo1=1, Hi1=n外环

1 to N

总计 Inner* Outer

5*N=5N

Answer 2

我认为最容易理解的方法是运行它。

例如对于 n=10，以下 java 代码在每次迭代中打印 i 和 j 的值：

 for (int i=0; i<=10; i++){
     for (int j=i-2; j<=i+2; j++)
         System.out.println("i = " + i + ", j= " + j);
 }

结果如下：

i = 0, j= -2
i = 0, j= -1
i = 0, j= 0
i = 0, j= 1
i = 0, j= 2
i = 1, j= -1
i = 1, j= 0
i = 1, j= 1
i = 1, j= 2
i = 1, j= 3
i = 2, j= 0
i = 2, j= 1
i = 2, j= 2
i = 2, j= 3
i = 2, j= 4
i = 3, j= 1
i = 3, j= 2
i = 3, j= 3
i = 3, j= 4
i = 3, j= 5
i = 4, j= 2
i = 4, j= 3
i = 4, j= 4
i = 4, j= 5
i = 4, j= 6
i = 5, j= 3
i = 5, j= 4
i = 5, j= 5
i = 5, j= 6
i = 5, j= 7
i = 6, j= 4
i = 6, j= 5
i = 6, j= 6
i = 6, j= 7
i = 6, j= 8
i = 7, j= 5
i = 7, j= 6
i = 7, j= 7
i = 7, j= 8
i = 7, j= 9
i = 8, j= 6
i = 8, j= 7
i = 8, j= 8
i = 8, j= 9
i = 8, j= 10
i = 9, j= 7
i = 9, j= 8
i = 9, j= 9
i = 9, j= 10
i = 9, j= 11
i = 10, j= 8
i = 10, j= 9
i = 10, j= 10
i = 10, j= 11
i = 10, j= 12

您现在可以看到 i 有 10 次迭代，并且对于每个 i，您有 j(from i-2 to i+2) 的 5 次迭代。因此总共 50 (5*N) 次迭代。

Answer 3

假设有一个 for 循环 Java 代码

for(int i=3;i<=6;i++)
{
    //Statements to be executed
}

循环会执行4次(i=3,4,5,6)

如何计算？

公式=上限-下限+1

所以在上面的循环中就是6-3+1=4(上限=6,下限=3)

在你的问题中上限=i+2

下限=i-2

内循环执行总次数=i+2-(i-2)+1=i+2-i+2+1=5

外层循环会执行n次所以只需要乘以n

因此答案变为 5*n=5n

嵌套循环执行

Nested loop execution

printing

algorithm

loops

for-loop