leetcode: Combination Sum

http://oj.leetcode.com/problems/combination-sum/

Given a set of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

The same repeated number may be chosen from C unlimited number of times.

Note:
All numbers (including target) will be positive integers.
Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
The solution set must not contain duplicate combinations.
For example, given candidate set 2,3,6,7 and target 7, 
A solution set is: 
[7] 
[2, 2, 3] 

思路：

显然是递归，注意两点：

1. 数字可以重复使用。
2. 结果里面不能有重复，比如[2, 2, 4, 5]的情况，第1个2用过以后循环直接跳到4。

class Solution {
public:
    void internalCombinationSum(vector<int> &candidates,
            int start,
            int sum,
            int target,
            vector<int> &combination,
            vector<vector<int> > &result) {
        int size = candidates.size();
        
        if (sum == target) {
            result.push_back(combination);
            
            return;
        }
        else if ((start >= size) || (sum > target)) {
            return;
        }

        for (int i = start; i < size; ) {
            int v = candidates[i];
            
            combination.push_back(v);
            internalCombinationSum(candidates, i, sum + v, target, combination, result);
            combination.pop_back();

            int j = i + 1;
            
            while (j < size) {
                if (candidates[i] == candidates[j]) {
                    ++j;
                }
                else {
                    break;
                }
            }
            
            i = j;
        }
    }
    
    vector<vector<int> > combinationSum(vector<int> &candidates, int target) {
        vector<vector<int> > result;
        vector<int> combination;
        
        sort(candidates.begin(), candidates.end());
        internalCombinationSum(candidates, 0, 0, target, combination, result);
           
        return result;
    }
};