Leetcode 40：Combination Sum II

原创于 2019-07-30 09:07:07 发布 · 135 阅读

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本文探讨了如何使用深度优先搜索(DFS)算法解决组合总和问题的变种，即在给定候选数集合和目标数的情况下，找出所有可能的唯一组合，使得组合中数的和等于目标数。特别强调了每个数在组合中只能使用一次，且解决方案必须排除重复组合。通过两个具体例子，详细展示了算法的实现过程。

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

Each number in candidates 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.
Example 1:

Input: candidates = [10,1,2,7,6,1,5], target = 8,
A solution set is:
[
[1, 7],
[1, 2, 5],
[2, 6],
[1, 1, 6]
]
Example 2:

Input: candidates = [2,5,2,1,2], target = 5,
A solution set is:
[
[1,2,2],
[5]
]

C++
 vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        std::sort(candidates.begin(),candidates.end());
        vector<vector<int>> res;
        vector<int> item;
        DFS(res,item,candidates,target,0);
        return res;
    }
    void DFS(vector<vector<int>>& res,vector<int> item,vector<int>& candidates,int sum,int n)
    {
        if(sum == 0)
            res.push_back(item);
        if(sum < 0)
            return;
        for(int i = n;i < candidates.size();++i)
        {
            if(i > n&&candidates[i] == candidates[i-1])continue;
            item.push_back(candidates[i]);
            DFS(res,item,candidates,sum - candidates[i],i+1);
            item.pop_back();
        }
    }