Combination Sum II

最新推荐文章于 2025-05-25 18:32:47 发布

转载最新推荐文章于 2025-05-25 18:32:47 发布 · 117 阅读

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原文链接：http://www.cnblogs.com/zhhwgis/p/3958221.html

本文探讨了在候选数字集合中寻找所有唯一组合的问题，这些组合的数字之和等于目标数值。特别强调了每个数字只能使用一次且结果需去除重复组合的要求。文中提供了一个具体的示例，并附带详细的C++实现代码。

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]

注意：结果需要排序，不能有相同的结果

 void dfs(vector<int> &sum, int nStart, int target, vector<int> tmp, vector<vector<int> > &res)
    {
        short nNum = sum.size();
        if (nStart > nNum)
            return;
    
        if (target < 0)
            return;
    
        if (target == 0)
        {
            res.push_back(tmp);
            return;
        }
    
        for (int i = nStart; i < sum.size(); i++)
        {
            if (sum[i] > target)
                continue;
            tmp.push_back(sum[i]);
            dfs(sum, i+1, target-sum[i], tmp, res);
            tmp.pop_back();
            while (i < sum.size() - 1 && sum[i] == sum[i+1])
                i++;
        }
    }
    
    vector<vector<int> > combinationSum2(vector<int> &num, int target) {
        vector<vector<int> > res;
        vector<int> tmp;
        sort(num.begin(), num.end());
        dfs(num, 0, target, tmp, res);
        sort(res.begin(),res.end());
        return res;
    }