如何将这种递归Java方法转换为迭代方法？（CodeJam挑战赛）

我正试图将这一点适应Python的代码阻塞问题。不幸的是，它需要太深的递归才能在Python中正常工作（除非递归限制和堆栈大小都显著增加）。所以我尝试将这个递归转换为迭代：

/**
 * Attempt to recursively free a die by selecting a different die for the same value.
 * @return true if the die has been freed, false if no other die can be found.
 */
boolean freeByShuffling(Die die) {
    assert die.valueUsing != null;
    // First check if we can just use another dice for the previous value
    for (Die otherDie : die.valueUsing.dice) {
        if (otherDie.valueUsing == null) {
            otherDie.valueUsing = die.valueUsing;
            die.valueUsing = null;
            return true;
        }
    }
    // Nope, we must free a die recursively
    diceVisitedWhileShuffling.add(die);
    for (Die otherDie : die.valueUsing.dice) {
        if (diceVisitedWhileShuffling.contains(otherDie)) continue;
        if (freeByShuffling(otherDie)) {
            otherDie.valueUsing = die.valueUsing;
            die.valueUsing = null;
            return true;
        }
    }
    return false;
}

这是我的Python代码，虽然它解决了大多数测试用例，但并不十分有效：

def free_by_shuffling(self, die):
    assert die.current_value is not None

    stack = [(None, die)]
    found = False

    while stack:
        this_die, other_die = stack.pop()
        self.visited.add(other_die)

        if found:
            other_die.current_value = this_die.current_value
            this_die.current_value = None
            continue

        for next_die in other_die.current_value.dice:
            if next_die in self.visited:
                continue
            if next_die.current_value is None:
                found = True
                stack.append((other_die, next_die))
                break
        else:
            for next_die in other_die.current_value.dice:
                if next_die in self.visited:
                    continue
                stack.append((other_die, next_die))

    return found

---

如何将原始方法转换为使用迭代而不是递归？

此Python实现适用于“小”和“大”输入文件：

def free_by_shuffling(self, die):
    assert die.current_value is not None

    stack = [die]
    found = False

    while stack:
        this_die = stack.pop()

        if found:
            if stack:
                this_die.current_value = stack[-1].current_value
                stack[-1].current_value = None
            continue

        for other_die in this_die.current_value.dice:
            if other_die.current_value is None:
                stack.extend((this_die, other_die))
                found = True
                break
        else:
            self.visited.add(this_die)

            for other_die in this_die.current_value.dice:
                if other_die not in self.visited:
                    stack.extend((this_die, other_die))
                    break
            else:
                if stack:
                    for other_die in stack[-1].current_value.dice:
                        if other_die not in self.visited:
                            stack.append(other_die)
                            break
    return found

---

欢迎您提出任何意见和建议。

您的解决方案“不太有效”的原因是什么？@Chris：少数测试用例的答案与迭代函数有误（通过

1

或

2

关闭），当使用递归的时候，所有的答案都是正确的。我在你的解决方案中改变了一些东西，现在只有4个测试用例是不正确的。我会继续试着看我是否能得到它。我被这个问题上的投票弄糊涂了。它是否明显违反了“寻求调试帮助的问题（“此代码为什么不工作？”）必须包括所需的行为、特定的问题或错误以及在问题本身中重现它所需的最短代码”？“不太管用”和“少数测试用例的错误率为1或2”（特别是没有细节）都不是明确的问题陈述。OP在这里已经有很长时间了；他们肯定知道这不是一个好问题吗？@Chris：再读一遍标题。我不寻求调试帮助。