leetcode 39 组合总和

组合总和

暴力回溯（无剪枝，时间复杂度高）

class Solution {
public:
    vector<vector<int>> result;
    vector<int> path;
    int sum;
    void backtracking(vector<int>& candidates, int target , int sum)
    {	
    	//检测目标大于时返回
        if(sum > target) return;
        if(sum == target)
        {
        	//排序后发现是新结果插入
            vector<int> tmp(path.begin(),path.end());
            sort(tmp.begin(),tmp.end());
            auto it = find(result.begin(),result.end(),tmp);
            if(it == result.end()) result.push_back(tmp);
            return;
        }
		//无任何限制回溯
        for(int i = 0 ; i < candidates.size() ;i++)
        {
            path.push_back(candidates[i]);
            backtracking(candidates,target,sum+candidates[i]);
            path.pop_back();
        }
        return;
    }
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
      
        backtracking(candidates,target,0);
        return result;
    }
};

回溯剪枝

class Solution {
public:
    vector<vector<int>> result;
    vector<int> path;
    int sum;
    void backtracking(vector<int>& candidates, int target , int sum , int indnx)
    {
        if(sum > target) return;
        if(sum == target)
        {
            result.push_back(path);
            return;
        }
  		//剪枝，因为之前已经对输入进行排序，当发现加上i点的值大于目标后，后面的也都大于
        for(int i = indnx ; i < candidates.size() && sum+candidates[i] <= target ;i++)
        {
            path.push_back(candidates[i]);
            //递归的下一个指针和当前一样都是i，不是i+1 
            //因为一个数可以重复的使用，不能重复是i+1
            backtracking(candidates,target,sum+candidates[i] , i);
            path.pop_back();
        }
        return;
    }
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        //对输入进行排序，方便后面循环
        sort(candidates.begin(),candidates.end());
        backtracking(candidates,target,0,0);
        return result;
    }
};

二刷

class Solution {
public:
    vector<int> path;
    vector<vector<int>> result;
    int sumNum=0;

    void dfs(vector<int>& candidates, int target,int fir)
    {
        if(sumNum > target) return;
        if(sumNum == target)
        {
            result.push_back(path);
            return;
        }

        for(int i=fir ; i<candidates.size() ; i++)
        {
            path.push_back(candidates[i]);
            sumNum += candidates[i];
            dfs(candidates,target,i);
            sumNum -= candidates[i];
            path.pop_back();
        }
    }

    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        dfs(candidates,target,0);
        return result;
    }
};