了解递归以生成排列

我发现递归，除了像阶乘这样非常直接的递归之外，非常难以理解。以下代码片段打印字符串的所有排列。谁能帮我理解一下。如何正确理解递归？

void permute(char a[], int i, int n)
{
   int j;
   if (i == n)
     cout << a << endl;
   else
   {
       for (j = i; j <= n; j++)
       {
          swap(a[i], a[j]);          
          permute(a, i+1, n);
          swap(a[i], a[j]);
       }
   }
} 

int main()
{
   char a[] = "ABCD";
   permute(a, 0, 3);
   getchar();
   return 0;
}

PaulR 有正确的建议。您必须“手动”运行代码（使用您想要的任何工具 - 调试器、纸张、在某些点记录函数调用和变量），直到您理解它。对于代码的解释，我将推荐您参考 quasiverse 的优秀答案。

Perhaps this visualization of the call graph with a slightly smaller string makes it more obvious how it works:

该图是用graphviz.

// x.dot
// dot x.dot -Tpng -o x.png
digraph x {
rankdir=LR
size="16,10"

node [label="permute(\"ABC\", 0, 2)"] n0;
 node [label="permute(\"ABC\", 1, 2)"] n1;
  node [label="permute(\"ABC\", 2, 2)"] n2;
  node [label="permute(\"ACB\", 2, 2)"] n3;
 node [label="permute(\"BAC\", 1, 2)"] n4;
  node [label="permute(\"BAC\", 2, 2)"] n5;
  node [label="permute(\"BCA\", 2, 2)"] n6;
 node [label="permute(\"CBA\", 1, 2)"] n7;
  node [label="permute(\"CBA\", 2, 2)"] n8;
  node [label="permute(\"CAB\", 2, 2)"] n9;

n0 -> n1 [label="swap(0, 0)"];
n0 -> n4 [label="swap(0, 1)"];
n0 -> n7 [label="swap(0, 2)"];

n1 -> n2 [label="swap(1, 1)"];
n1 -> n3 [label="swap(1, 2)"];

n4 -> n5 [label="swap(1, 1)"];
n4 -> n6 [label="swap(1, 2)"];

n7 -> n8 [label="swap(1, 1)"];
n7 -> n9 [label="swap(1, 2)"];
}