将递归置换生成器转换为迭代

Question

我在转换这个递归算法时遇到了一些困难,该算法用于将给定整数集的所有排列显示为迭代整数.

void getPermutationsR(int v[], int n, int i) 
{
    if (i == n)
    {
        //Display contents of v
    } 
    else
    {
        for (int j=i; j<n; j++) 
        {
            swap (v, i, j);
            getPermutationsR(v, n, i+1);
            swap (v, i, j);
        }
    }
}

这是我目前的尝试,这是完全错误的,但我没有看到任何方法来纠正它,而不使用本机迭代算法的问题.我的一半尝试让我'弹出'超过'推'(当我试图访问空堆栈中的元素时导致错误)而另一半我'推''超过'弹出'(无限循环).

void getPermutationsI(int v[], int n, int i) 
    {
    stack<int> iStack;
    stack<int> jStack;

    iStack.push(i);
    jStack.push(i);

    while(iStack.size() > 0)
    {
        if (iStack.top() == n)
        {
            jStack.pop();
            iStack.pop();
            //Display contents of v
        }
        else
        {
            for (int j = iStack.top(); j < n; j++)
            {
               //swap 
                               //something to do with jStack
            }
            //swap 
            iStack.push(i+1);
        }
    }
}

Answer 1

您遇到的挑战是您已经混合了函数调用和循环结构.很难解开那些.

首先让我们开始用递归替换所有流操作控制.

// You'll want to start with getPermutionsR(v, n, 0, 0)
void getPermutationsR(int v[], int n, int i, int j) 
{
    if (i == n)
    {
        //Display contents of v
    }
    else if (j == n) {
        // By doing nothing, we break out of the loop
    }
    else
    {
        // This was your recursive call inside of the loop.
        // Note that I'm sending you to to the first loop iteration here.
        swap (v, i, j);
        getPermutationsR(v, n, i+1, i+1);
        swap (v, i, j);

        // And the next loop iteration
        getPermutationsR(v, n, i, j+1);
    }
}

接下来让我们添加更多状态,这样我们在if条件中只有一个递归调用.

// You'll want to start with getPermutionsR(v, n, 0, 0, 1)
void getPermutationsR(int v[], int n, int i, int j, bool firstCall)
{
    if (i == n)
    {
        //Display contents of v
    }

    int x = i;
    int y = j+1;
    if (firstCall) {
        swap (v, i, j);
        x = i+1;
        y = i+1;
    }

    // My one recursive call.  Note that i=n implies j=n.
    if (j < n) {
        getPermutationsR(v, n, x, y, !firstCall);
    }

    if (firstCall) {
        swap (v, i, j);
    }
}

现在我们已经完成了这个,我们将它以一种形式,我们可以以一种简单的方式将其转换为迭代版本.这是转型

void recursiveForm (params, recursiveState) {
    topHalf...
    if (cond) {
        recursiveForm(...)
    }
    bottomHalf...
}

变

void iterativeForm(params) {
    initializeStacks...
    pushStacks...
    topHalf...
    while (stacks not empty) {
        if (cond) {
            pushStacks...
            topHalf...
        }
        else {
            bottomHalf...
            popStacks...
        }
    }
}

所以应用这种模式我们得到类似的东西:

// You'll want to start with getPermutionsR(v, n, 0, 0, 1)
void getPermutationsI(int v[], int n)
{
    stack<int> iStack;
    stack<int> jStack;
    stack<bool> firstCallStack;

    // pushStacks with initial value
    iStack.push(0);
    jStack.push(0);
    firstCallStack.push(1);

    // topHalf...
    if (iStack.top() == n)
    {
        //Display contents of v
    }

    int x = iStack.top();
    int y = jStack.top()+1;
    if (firstCallStack.top()) {
        swap (v, iStack.top(), jStack.top());
        x = iStack.top()+1;
        y = iStack.top()+1;
    }

    while iStack.size() > 0) {
        // if (cond) {
        if (jStack.top() < n) {
            //pushStacks...
            iStack.push(x);
            jStack.push(y);
            firstCallStack.push(!firstCallStack.top());

            // topHalf...
            if (iStack.top() == n)
            {
                //Display contents of v
            }

            x = iStack.top();
            y = jStack.top()+1;
            if (firstCallStack.top()) {
                swap (v, iStack.top(), jStack.top());
                x = iStack.top()+1;
                y = iStack.top()+1;
            }
        }
        else {
            // bottomHalf...
            if (firstCallStack.top()) {
                swap (v, iStack.top(), jStack.top());
            }
        }
    }
}

(警告,所有代码都未经测试,甚至可能无法编译.但这个想法是正确的.它绝对是相同的算法.)