在编写C#时遇到了这个问题,但我认为答案适用于任何编程语言.
我需要多个嵌套循环(让我们说9,但可以是100).我的代码应该执行如下:
for (a1=0;a1<N;a1++)
for (a2=0;a2<N;a2++)
...
for (a9=0;a9<N;a9++)
//do something
Run Code Online (Sandbox Code Playgroud)
我认为通过使用变量M=9
和int数组a[]
而不是a1,a2,a3,... 可以缩短此代码的速度.
你能提供任何提示吗?
编辑:
我知道我不需要使用嵌套循环来执行此操作.我只是问我是否可以使用变量M=9
int数组a[]
和更少的代码来做到这一点.
如果你真的需要看一些复杂的东西,我写了下面的代码:
string[] s1 = new string[3]{ "apples", "oranges","bananas" };
string[] s2 = new string[5] { "are", "are not", "should be", "must be","seem to be" };
string[] s3 = new string[3] { "big", "small", "tasty" };
int[] a=new int[10];
int[] step = new int[10];
step[1] = 1; step[2] = 1; step[3] = 2;
for (a[1] = 0; a[1] < s1.Length; a[1]+=step[1])
for (a[2] = 0; a[2] < s2.Length; a[2]+=step[2])
for (a[3] = 0; a[3] < s3.Length; a[3]+=step[3])
Console.WriteLine(s1[a[1]] + " " + s2[a[2]] + " " + s3[a[3]]);
Run Code Online (Sandbox Code Playgroud)
想象一下有更多的数组:s4,s5,... s10.(这可以是数组数组s[]
,也可以是二维数组s[,]
.
limits数组定义每个组件的上限(-1,相当于for循环中的<N).components数组包含每个组件的值.
int[] limits = { 9, 20, 15, 30, 8, 40, 10 };
int[] components = new int[limits.Length];
bool done = false;
while (!done)
{
//do work here using the components
int M = components[0];
int N = components[1];
int O = components[2];
int P = components[3];
//etc
//first iteration M=0,N=0,O=0, etc
//this part does the incrementing for the next iteration
for (int j = 0; j < limits.Length; j++)
{
components[j]++;
if (components[j] < limits[j])
break;
components[j] = 0;
if (j == limits.Length - 1)
done = true;
}
}
Run Code Online (Sandbox Code Playgroud)