LeetCode·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： Informal essay

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本文探讨了组合总和问题，即给定一组候选数和目标数，找出所有可能的组合，使得候选数之和等于目标数。解决方案采用了递归算法，并提供了详细的C++实现代码示例。

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题目：

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

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

Note:

All numbers (including target) will be positive integers.
The solution set must not contain duplicate combinations.
Example 1:

Input: candidates = [2,3,6,7], target = 7,
A solution set is:
[
  [7],
  [2,2,3]
]
Example 2:

Input: candidates = [2,3,5], target = 8,
A solution set is:
[
  [2,2,2,2],
  [2,3,3],
  [3,5]
]

思路：

递归

代码：

class Solution {
public:
    void help(vector<int>& nums, vector<vector<int> > &vec, vector<int> &v, int start, int target){
        if(target < 0)return;
        if(target == 0){
            vec.push_back(v);
            return;
        }

        for(int i = start; i < nums.size(); ++i){
            v.push_back(nums[i]);
            help(nums, vec, v, i, target-nums[i]);
            v.pop_back();
        }
    }

    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        vector<vector<int> > vec;
        vector<int> v;
        if(candidates.size() < 1)return vec;
        //题中讲无重复数字，所以可以不排序
        //如果排序，可以进行提前终止，只处理小于target的数字
        //sort(candidates.begin(), candidates.end());
        help(candidates, vec, v, 0, target);
        return vec;
    }
};

结果：