使用递归将所有可能的组合（带有重复）作为数组中的值

Question

I'm trying to solve a problem in which I need to insert math operations(+/- in this case) between digits or merge them to get a requested number.

For ex.: 123456789 => 123+4-5+6-7+8-9 = 120

My concept is basically generating different combinations of operation codes in array and calculating the expression until it equals some number.

The problem is I can't think of a way to generate every possible combination of math operations using recursion.

Here's the code:

#include <iostream>
#include <algorithm>

using namespace std;

enum {noop,opplus,opminus};//opcodes: 0,1,2

int applyOp(int opcode,int x, int y);
int calculate(int *digits,int *opcodes, int length);
void nextCombination();

int main()
{
    int digits[9] = {1,2,3,4,5,6,7,8,9};
    int wantedNumber = 100;

    int length = sizeof(digits)/sizeof(digits[0]);

    int opcodes[length-1];//math symbols
    fill_n(opcodes,length-1,0);//init

    while(calculate(digits,opcodes,length) != wantedNumber)
    {
        //recursive combination function here
    }

    return 0;
}
int applyOp(int opcode,int x, int y)
{
    int result = x;
    switch(opcode)
    {
        case noop://merge 2 digits together
            result = x*10 + y;
            break;
        case opminus:
            result -= y;
            break;
        case opplus:
        default:
            result += y;
            break;
    }
    return result;
}
int calculate(int *digits,int *opcodes, int length)
{
    int result = digits[0];
    for(int i = 0;i < length-1; ++i)//elem count
    {
        result = applyOp(opcodes[i],result,digits[i+1]);//left to right, no priority
    }
    return result;
}

Answer 1

The key is backtracking. Each level of recursion handles a single digit; in addition, you'll want to stop the recursion one you've finished.

The simplest way to do this is to define a Solver class, which keeps track of the global information, like the generated string so far and the running total, and make the recursive function a member. Basically something like:

class Solver
{
    std::string const input;
    int const target;

    std::string solution;
    int total;
    bool isSolved;

    void doSolve( std::string::const_iterator pos );
public:
    Solver( std::string const& input, int target )
        : input( input )
        , target( target )
    {
    }

    std::string solve()
    {
        total = 0;
        isSolved = false;
        doSolve( input.begin() );
        return isSolved
            ? solution
            : "no solution found";
    }
};

In doSolve , you'll have to first check whether you've finished ( pos == input.end() ): if so, set isSolved = total == target and return immediately; otherwise, try the three possibilities, ( total = 10 * total + toDigit(*pos) , total += toDigit(*pos) , and total -= toDigit(*pos) ), each time saving the original total and solution , adding the necessary text to solution , and calling doSolve with the incremented pos . On returning from the recursive call, if ! isSolved ! isSolved , restore the previous values of total and solution , and try the next possibility. Return as soon as you see isSolved , or when all three possibilities have been solved.