leetcode 40 Combination Sum II

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

Each number in candidates may only be used once in the combination.

Note:

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

Example 1:

Input: candidates = [10,1,2,7,6,1,5], target = 8,
A solution set is:
[
  [1, 7],
  [1, 2, 5],
  [2, 6],
  [1, 1, 6]
]

继上一篇博客（leetcode 90 Subsets II）写道，求子集、符合条件的情况的题目一般都是可以用回溯法来解决的。本题也是，需注意到是本题有重复元素，在结果中会产生重复的子集，需要在加入时进行手动剔除。同时，和第39题（Combination Sum）有区别的是，本题一个元素只能使用一次，因此在回溯时需要对索引+1。代码记录如下：

class Solution {
public:
    vector<vector<int>> combinationSum(vector<int>& candidates, int target){
        vector<vector<int>> ret;
        vector<int> tmp;
        ret.clear();
        tmp.clear();
    
    
        sort(candidates.begin(), candidates.end());
        backtrack(candidates, ret, tmp, target, 0);
        
        return ret;
    }
        
    void backtrack(const vector<int> &candidates, vector<vector<int>> &ret, 
                   vector<int> &tmp, int target, int idx)
    {
        if(0 == target)
        {
            ret.push_back(tmp);
            return;
        }
        else
        {
            for(int i = idx; i < candidates.size(); ++i)
            {
                if(candidates[i] > target) 
                    return;
                tmp.push_back(candidates[i]);
                backtrack(candidates, ret, tmp, target - candidates[i], i); //Attension: i is here!
                tmp.pop_back();
            }
        }
    }

   
};