嵌套循环执行

Question

这是一个非常简单的问题，但由于某种原因，我感到困惑，这很烦人。

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

body() for语句执行多少次?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 =我2,Hi2 =我+ 2 内循环

所以(i-2)到(i+2)=5变式EG: i-2 , i-1 , i , i+1 , i+2

Like (-2) to (+2)=-2，-1,0,1,2

Lo1=1, Hi1=n Outer Loop

1到N

所以Total Inner* Outer

5 * N = 5 N

Answer 2

我认为最简单的理解方法就是运行它。

以n=10为例，下面的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迭代，而对于每个i，您有j(from i-2 to i+2)的5迭代。因此总共有50 (5*N)次迭代

Answer 3

假设Java代码中有一个for循环

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

循环将执行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，它将是6-3+1=4(上界=6，下界=3)

在你的问题上限=我+ 2

下界=我2

内循环执行的总次数=i+2-(i-2)+1=i+2-i+2+1=5

外部循环将执行n次所以只需乘以n

因此答案变成5*n=5n