java - 坚持理解递归

Question

在过去的几个小时里，我试图理解为什么这个程序的输出表现得如此奇怪。递归函数的第一部分对我来说很有意义，但是不幸的是，我发现其余部分非常困惑。

我想知道为什么程序会倒退，甚至添加到一个应该被减去的变量。这对我来说没有任何意义。如果有任何帮助，我将不胜感激。

这是我的递归函数:

public static void threeWays(int id, int n) {
    if (n >= 0) {
        System.out.println(id + "=" + n);
        threeWays(id + 1, n - 1);
        threeWays(id + 1, n - 2);
        threeWays(id + 1, n - 3);
    }
}

public static void main(String[] args) {
    threeWays(1, 4);
}

这是输出:

1=4

2=3

3=2

4=1

5=0

4=0

3=1

4=0

3=0

2=2

3=1

4=0

3=0

2=1

3=0

Answer 1

Python3版本:

  def threeWays(f, i, n):
      if n >= 0:
          print(f, i, '=', n)
          threeWays('1st', i + 1, n - 1)
          threeWays('2nd', i + 1, n - 2)
          threeWays('3rd', i + 1, n - 3)

  threeWays('    ', 1, 4)

乍一看，它可以，可以看起来像DFS但实际上不是...

当 threeWays 从 2nd 和 3rd threeWays 递归调用时，第 1 或 1st 和 2nd threeWay 也会被调用。 ..!

threeWays('   ', 1, 4)            ->     1 = 4
  threeWays('1st', 2, 3)          -> 1st 2 = 3
    threeWays('1st', 3, 2)        -> 1st 3 = 2
      threeWays('1st', 4, 1)      -> 1st 4 = 1
        threeWays('1st', 5, 0)    -> 1st 5 = 0
          threeWays('1st', 6, -1)
        threeWays('2nd', 5, -2)
        threeWays('3rd', 5, -3)
      threeWays('2nd', 4, 0)      -> 2nd 4 = 0
      threeWays('3rd', 4, -1)
    threeWays('2nd', 3, 1)        -> 2nd 3 = 1
      threeWays('1st', 4, 0)      -> 1st 4 = 0
      threeWays('2nd', 4, -1)
      threeWays('3rd', 4, -2)
    threeWays('3rd', 3, 0)        -> 3rd 3 = 0
...