LeetCode 39. Combination Sum

Posted 2020-06-16

Given a set of candidate numbers (C) 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.
• Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
• 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 getCombinationSum(int currentSum,vector<int>& candidates, int index,vector <vector<int>> &solution, vector <int> &result, int target){
        if(currentSum == target){ //如果和等于目标，则把该结果压入solution，返回上一          层
            solution.push_back(result);
            return;
        }
        else{  //如果和小于目标，则可以继续加入值
            for(int i = index; i < candidates.size() ;i++){
                currentSum += candidates[i];
                if(currentSum> target){//加值后大于目标，则该队列不可能再复合要求，直接break该循环
                  currentSum -= candidates[i];
                  break;
                }
                result.push_back(candidates[i]);//把当前结果压入，继续递归调用
                getCombinationSum(currentSum,candidates,i,solution,result,target);
                currentSum -= candidates[i];//返回的情况说明之前找到了一个解，所以把最后压入的去掉，然后继续循环添加新值
                result.pop_back();
            }
        }
        
        
    }

   
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        vector <vector<int>> solution;
        vector <int> result;
        int currentSum = 0;
        sort(candidates.begin(),candidates.end());//先把输入排序，确保输出是有序的
        getCombinationSum(currentSum,candidates,0,solution,result,target);
        return solution;
    }
};