leetcode:Combination Sum II

最新推荐文章于 2019-03-06 12:45:47 发布

martin_liang

最新推荐文章于 2019-03-06 12:45:47 发布

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本文详细解析了一种组合求和算法，该算法用于从候选数集合中找出所有可能的组合，使得这些组合中的数相加等于目标数。文章强调了算法的实现细节，包括去除重复组合、确保组合内元素递增排列等关键步骤。

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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.
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 10,1,2,7,6,1,5 and target 8,
A solution set is:
[1, 7]
[1, 2, 5]
[2, 6]
[1, 1, 6]

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class Solution {
    
private:
    void combination(vector<int>& candidates, int idx, int target, vector<int>& curVtr, vector<vector<int>>& retVtr)
    {
        if (target == 0)
        {
            retVtr.push_back(curVtr);
            return;
        }
        
        for (int i=idx; i<candidates.size(); i++)
        {
            if(candidates[i]>target)
                return;
                
            if (i>idx && candidates[i] == candidates[i-1])
                continue;
            
            curVtr.push_back(candidates[i]);
            
            combination(candidates, i+1, target-candidates[i], curVtr, retVtr);
            
            curVtr.pop_back();
        }
    }
    
public:
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        
        vector<vector<int> > retVtr;
        vector<int> curVtr;
        
        sort(candidates.begin(), candidates.end());
        combination(candidates, 0, target, curVtr, retVtr);
        
        return retVtr;
        
    }
};