40. Combination Sum II

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本文探讨了给定一组候选数和目标数时如何找出所有不重复的组合，使得候选数之和等于目标数。候选数在每种组合中只能使用一次。通过C++代码实现了一种解决方案，并详细介绍了其工作原理。

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

Each number in C 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.

For example, given candidate set [10, 1, 2, 7, 6, 1, 5] and target 8,

A solution set is:

[
  [1, 7],
  [1, 2, 5],
  [2, 6],
  [1, 1, 6]
]

class Solution {
public:
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        sort(candidates.begin(), candidates.end());
        vector<vector<int>> res;
        vector<int> combination;
        combinationSum2(candidates, target, res, combination, 0);
        return res;
        
    }
    
private:
    void combinationSum2(vector<int>& candidates, int target, vector<vector<int>>& res, vector<int>& combination, int begin){
        if(!target){
            res.push_back(combination);
            return;
        }
        for(int i = begin; i < candidates.size() && target >= candidates[i]; ++i){
                if(i > begin && candidates[i] == candidates[i-1]) continue; //关键在于是递归进去的还是 i自加进去，后者要去除
                combination.push_back(candidates[i]);
                combinationSum2(candidates, target-candidates[i], res, combination, i+1);
                combination.pop_back();
           
            
        }
        
    }
};