[Leetcode]Combination Sum II

最新推荐文章于 2019-05-10 10:55:45 发布

原创最新推荐文章于 2019-05-10 10:55:45 发布 · 231 阅读

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本文详细介绍了如何使用深度优先搜索（DFS）解决组合总和 II 问题，即从给定的候选数集合中找到所有唯一的组合，使得这些数的和等于给定的目标值。文章涵盖了算法实现细节、注意事项以及复杂度分析。

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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:
    void dfs(vector<vector<int> >&result,vector<int>&path,vector<int>& nums,int target,int start)
    {
        if(target < 0)return;
        if(target == 0){ //get one solution
            result.push_back(path);
            return;
        }
        for(int i = start;i < nums.size();i++){
            path.push_back(nums[i]);
            dfs(result,path,nums,target - nums[i],i+1);
            path.pop_back();
            while(i+1<nums.size()&&nums[i]==nums[i+1])++i;//skip duplicate
        }
    }
    //[1,1,2,5,6,7,10]
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
            vector<vector<int>>result;
            vector<int>path;
            sort(candidates.begin(),candidates.end());
            dfs(result,path,candidates,target,0);
            return result;
    }
};