LeetCode(39) Combination Sum

题目如下：

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] 

分析如下：

写了几道DFS的题目之后， 发现这道题目就是标准的DFS题目，具体分下见下面的代码。

我的代码：

class Solution {
public:
    void combinationSumMy(vector<int> &candidates, vector<int> &every, vector<vector<int> > & final, int target, int start) {
        if (target == 0) { //说明已经找到了合适的结果向量，保存当前的结果向量every到2维向量final中
            final.push_back(every);
            return;
        }else {
            for (int i = start; i < candidates.size(); ++i) {
                if (candidates[i] > target) return; //剪枝，因为题目说了都是正数
                every.push_back(candidates[i]); //把当前candidate[i]放入结果向量中, target 变化为target - candidate[i]
                combinationSumMy(candidates, every, final, target - candidates[i], i); //对新target(target - candidate[i]) 进行DFS
                every.pop_back(); //回溯
            }
        }
    }
    vector<vector<int> > combinationSum(vector<int> &candidates, int target) {
        std::sort(candidates.begin(), candidates.end());
        vector<int> every;
        vector<vector<int> > final;
        combinationSumMy(candidates, every, final, target, 0);
        return final;
    }
};

一开始写得版本是错误的，没有滤重，对比下面这个错误版本，可以更清晰第认识到start起到的作用。start使得每次选到的candidate[i]都只能为两种情况，要么是start本身， 要么是start右边的数。

//错误版本
class Solution {
public:
    void combinationSumMy(vector<int> &candidates, vector<int> &every, vector<vector<int> > & final, int target) {
        if (target < 0) {
            return;
        }else if (target == 0) {
            final.push_back(every);
            return;
        }else {
            for (int i = 0; i < candidates.size(); ++i) {
                every.push_back(candidates[i]);
                combinationSumMy(candidates, every, final, target - candidates[i]);
                every.pop_back();
            }
        }
    }
    vector<vector<int> > combinationSum(vector<int> &candidates, int target) {
        std::sort(candidates.begin(), candidates.end());
        vector<int> every;
        vector<vector<int> > final;
        combinationSumMy(candidates, every, final, target);
        return final;
    }
};

错误的输出

2,2,3,
2,3,2,
3,2,2,
7,

因为输出有重复，所以增加了start,用来避免重复输出，每次的candidate都是start本身或者start右边的数。