【LeetCode 53】39.组合总和

【LeetCode 53】39.组合总和

一、题意

二、解答过程

回溯三部曲:

  • 确定参数
  • 确定终止条件
  • 确定单层递归逻辑
class Solution {
    
    
private:
    vector<vector<int>>result;
    vector<int> path;
    void backtracking(vector<int>&candidates,int target,int sum,int startIndex)
    {
    
    
        if(sum>target)
        {
    
    
            return;
        }
        if(sum==target)
        {
    
    
            result.push_back(path);
            return;
        }

        
          for (int i = startIndex; i < candidates.size(); i++)
         //剪枝操作
        //for(int i=startIndex;i<candidates.size()&&sum+candidates[i]<=target;i++)
        {
    
    
            /* code */
            sum+=candidates[i];
            path.push_back(candidates[i]);
            //不需要i+1表示可以重复读取当前的数
            backtracking(candidates,target,sum,i);
            sum-=candidates[i];
            path.pop_back();
        }
        

    }
public:
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
    
    
        result.clear();
        path.clear();
        backtracking(candidates,target,0,0);
        return result;
    }
};

剪枝操作的理解,就是for循环那里做改动:

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转载自blog.csdn.net/weixin_43891901/article/details/123033751