combination sum

Given a set of candidate numbers (candidates) (without duplicates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.

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

Note:

• All numbers (including target) will be positive integers.
• The solution set must not contain duplicate combinations.

Example 1:

Input: candidates = [2,3,6,7], target = 7,
A solution set is:
[
  [7],
  [2,2,3]
]

Example 2:

Input: candidates = [2,3,5], target = 8,
A solution set is:
[
  [2,2,2,2],
  [2,3,3],
  [3,5]
]

我的解答：

class Solution {

public:
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        vector<vector<int>> res;
        vector<int> temp;
        //if (candidates.size() < 1)return res;
        sort(candidates.begin(),candidates.end());
        combinationSum(candidates,res,temp,target,0);
        return res;
    }

private:
    void combinationSum(vector<int>& candidates, vector<vector<int>> &res,vector<int> &temp, int target,int begin)
    {
        if (!target)
        {
            res.push_back(temp);
            return ;
        }
        // 这里要注意了，&& 两边应该先判断 i ，再判断candidates[i],否则将导致数组越界！！！
        for (int i = begin; i != candidates.size() && target >= candidates[i]; ++i)
        {
            temp.push_back(candidates[i]);
            combinationSum(candidates,res,temp,target - candidates[i],i);
            temp.pop_back();
        }
    }
};

其中有一个要注意的地方：

 && 运算符：先判断左边，只有左边为真，才计算右边

 || 运算符：先判断左边，只有左边为假，才计算右边