Description

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]
]

分析

题目的意思是：给你一些候选的数，然后给出所有可能的组合数。

• 这是一个DFS经典题目，这是一个经典的解法，递归，注意终止条件就行了。

代码

class Solution {
public:
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        vector<vector<int>>  result;
        vector<int> ans;
        getSum(candidates,target,0,0,result,ans);
        return result;
    }
    
    void getSum(vector<int>& candidates, int target,int sum,int index,vector<vector<int>> &result,vector<int> ans){
        if(sum>target){
            return;
        }
        if(sum==target){
            result.push_back(ans);
        }
        for(int i=index;i<candidates.size();i++){
            ans.push_back(candidates[i]);
            getSum(candidates,target,sum+candidates[i],i,result,ans);
            ans.pop_back();
        }
    }
};

参考文献

​​[编程题]combination-sum​​