一个递归转循环的模式

有些递归是很容易转化成循环的，用一个循环非常直观的映射过去就是了，如求Fibonacci数列；而有些递归却没有那么直观，甚至可能需要多个循环才能转化过去，这里举个例子：

给出一个集合，如(1, 2, 3, 4)，打印出该集合的所有子集

分析一下问题，子集是指取原集合中的任意多个元素，转化一下问题，就是对于原集合中的任何一个元素，我们都有两个选择，包含或者不包含，所以对于n个元素的集合，其子集数为： 2*2*2... = 2^n。那么可以得出其递归算法：

void Recursive_Subsets(int* a, bool* b, int start, int end)
{
    if (start <= end)
    {
        // determine current number's status, which has 2 choices
        // and then continue with left ones
        b[start] = true;
        Recursive_Subsets(a, b, start+1, end);
 
        b[start] = false;
        Recursive_Subsets(a, b, start+1, end);
    }
    else
    {
        for (int i = 0; i <= end; i++)
        {
            if (b[i]) cout << a[i];
        }
        cout << endl;
    }
 
}
 
void PrintAllSubsets1(int* a, int n)
{
    bool* b = new bool[n];
    Recursive_Subsets(a, b, 0, n-1);
    delete b;}

递归的优点是直观、易懂：写起来如此，读起来也是这样。但是每次递归都是call stack的不断叠加，对于这个问题，其需要消耗O(n)的栈空间，栈空间，栈空间~~~

于是，我们也可以将其转化成循环的方式来实现：

void PrintAllSubsets2(int* a, int n)
{
    // Initialize flags as false
    bool* b = new bool[n];
    for(int i = 0; i < n; i++)
    {
        b[i] = false;
    }
 
    // Use 2 loops to enumerate all possible combinations of (each item's status * number of items), 
    // in this case: ([true|false] * size of set)
    while(true)
    {
        // Get one solution, output it!
        for(int i = 0; i < n; i++)
        {
            if(b[i]) cout << a[i];
        }
        cout << endl;
 
 
        // One of the number's status has switched, start over enumeration for that!
        // ex: I have following enumeration for 3 numbers:
        //     0, 0, 0
        //     0, 0, 1
        //     0, 1, 0
        //     0, 1, 1
        // now if I changed the first number's status from 0 to 1, I need to re-enumerate the other 2 numbers
        // to get all possible cases:
        //     1, 0, 0
        //     1, 0, 1
        //     1, 1, 0
        //     1, 1, 1
        int k = n - 1; 
 
        while(k >= 0)
        {
            if(b[k] == false)   // have we tried all possible status of k?
            {
                b[k] = true;    // status switch, as we only have 2 status here, I use a boolean rather than an array.
                break;          // break to output the updated status, and further enumeration for this status change, just like a recursion
            }
            else                // We have tried all possible cases for k-th number, now let's consider the previous one
            {
                b[k] = false;   // resume k to its initial status
                k--;            // let's consider k-1
            }
        }
 
        if(k < 0) break;     // all the numbers in the set has been processed, we are done!
    }
 
    // clean up
    delete [] b;
}