题目描述

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.

样例

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]
]

算法1

(回溯法） $O(2^n)$

C++ 代码

class Solution {
public:
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        vector<vector<int>> res;
        vector<int> vec;
        dfs(candidates,0,0,target,vec,res);
        return res;
    }
    //因为需保证最后组合结果唯一,dfs必须从n开始，如果之前的搜过，不能往上继续搜，所以这里有个n从n往下搜
    void dfs(vector<int>& candidates,int cur,int n,int target,vector<int>&vec,vector<vector<int>>&res){
        if(cur>=target){//当前结果大于target返回
            if(cur==target)
                res.push_back(vec);
            return;
        }
        for(int i=n;i<candidates.size();i++){
            vec.push_back(candidates[i]);
            dfs(candidates,cur+candidates[i],i,target,vec,res);
            vec.pop_back();
        }
    }
};