[LeetCode] #38 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] 

本题利用回溯算法求解，重点在于消除重复。时间：12ms。代码如下：

class Solution {
public:
    void makeCombination(vector<vector<int> > &ans, vector<int> &candidates, vector<int> &tmp, int target, int cur){
        for (int i = cur; i < candidates.size(); ++i){
            if (0 == target) {
                ans.push_back(tmp);
                return;
            }
            if (candidates[i] <= target){
                tmp.push_back(candidates[i]);
                makeCombination(ans, candidates, tmp, target - candidates[i], i);
                tmp.pop_back();
            }
            if (candidates[i] > target){
                return;
            }
        }
    }
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        vector< vector<int> > ret;
        sort(candidates.begin(), candidates.end());
        if (candidates.empty() || target < candidates[0])
            return ret;
        vector<int> v;
        makeCombination(ret, candidates, v, target, 0);
        return ret;    
    }
};