[LeetCode] Combination Sum II

最新推荐文章于 2019-05-07 10:37:18 发布

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本文介绍了一种寻找候选数集合中所有唯一组合的算法，这些组合的和等于目标数。文章详细解释了如何确保组合的独特性及非递减顺序，并通过实例展示了算法的有效性。

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]

class Solution {
public:
    vector<vector<int> >ans;
    set<vector<int> > res;
    vector<int> v;
    vector<vector<int> > combinationSum2(vector<int> &candidates, int target) {
        sort(candidates.begin(),candidates.end());
        generate(candidates,0,0,target);
        for(set<vector<int> >::iterator it = res.begin();it != res.end();it ++)
            ans.push_back(*it);
        return ans;
    }
    void generate(vector<int> cand,int sum,int start,int target){
        if(sum == target){
            res.insert(v);
            return;
        }
        for(int i = start;i < cand.size();++ i){
            v.push_back(cand[i]);
            sum += cand[i];
            if(sum > target){
                v.pop_back();
                break;
            }
            generate(cand,sum,i + 1,target);
            v.pop_back();
            sum -= cand[i];
        }
    }
};