39. Combination Sum

最新推荐文章于 2025-05-25 18:32:47 发布

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最新推荐文章于 2025-05-25 18:32:47 发布

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分类专栏： Leetcode

本文链接：https://blog.youkuaiyun.com/gcola007/article/details/76861400

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本文详细解析了给定一组候选数和目标数时如何找出所有可能的组合使得候选数之和等于目标数的问题。通过递归算法实现了重复利用候选数进行组合的解决方案，并确保了组合的唯一性。

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题目：

Given a set of candidate numbers (C) (without duplicates) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

The same repeated number may be chosen from C unlimited number of times.

Note:

All numbers (including target) will be positive integers.
The solution set must not contain duplicate combinations.

For example, given candidate set [2, 3, 6, 7] and target 7,
A solution set is:

[
  [7],
  [2, 2, 3]
]

class Solution {
public:
    void combinationSum1(vector<int>& candidates,int target,int begin,vector<vector<int>> &res,vector<int>& v){
        if(!target) {
            res.push_back(v);
            return;}
        for(int i=begin;i<candidates.size()&&target>=candidates[i];i++){
            v.push_back(candidates[i]);
            combinationSum1(candidates,target-candidates[i],i,res,v);
            v.pop_back();
        }
    }

    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        sort(candidates.begin(),candidates.end());
        vector<vector<int>> res;
        vector<int> v;
        combinationSum1(candidates,target,0,res,v);
        return res;
    }
};

遇见类似于求子集，或者选择与不选择

注意push_back与pop_back的灵活使用