Combination Sum II（c++实现）

最新推荐文章于 2025-08-06 00:28:11 发布

最新推荐文章于 2025-08-06 00:28:11 发布 · 178 阅读

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本文介绍了一种算法，用于在一个整数集合中查找所有唯一组合，这些组合的元素相加等于给定的目标值。算法确保每个候选数只使用一次，并且组合按非递减顺序排列，避免重复结果。

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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, � , a_k) must be in non-descending order. (ie, a₁ ? a₂ ? � ? a_k).
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> &num, int target) {
        vector<vector<int> > ret;
        if(target <= 0) return ret;
        if(num.size() == 0) return ret;
        sort(num.begin(), num.end());
        vector<int> tmp;
        iter(num, target, 0, tmp, ret);
        return ret;
    }
    
    void iter(vector<int> &num, int target, int beg, vector<int> tmp, vector<vector<int> > &ret) {
        if(target == 0) {
            ret.push_back(tmp);
            return;
        }
        if(target > 0) {
            for(int i = beg; i < num.size(); ++i) {
                if(i == beg || num[i] != num[i-1]) {
                    tmp.push_back(num[i]);
                    iter(num, target-num[i], i+1, tmp, ret);
                    tmp.pop_back();
                }
            }
        }
    }
    
};

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