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Leetcode_Combination Sum 回溯法

程序员文章站 2022-05-21 22:39:01
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39. Combination Sum

class Solution {
public:
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        sort(candidates.begin(),candidates.end());
        vector<vector<int>> res;
        vector<int> temp;
        backTrack(candidates,target,res,temp,0);
        return res;
    }
    void backTrack(vector<int>& candidates,int target,vector<vector<int> >& res,vector<int> &temp,int begin)
    {
        if(target==0)
        {
            res.push_back(temp);
            return;
        }
        for(int i=begin;i<candidates.size()&&candidates[i]<=target;i++)
        {
            temp.push_back(candidates[i]);
            backTrack(candidates,target-candidates[i],res,temp,i);
            temp.pop_back();
        }
    }
};


本题给出一个数组,该数组中的数字可重复使用,构造相加可以等于某个特定的数组,求这些数组的集合。

思路就是先排序,然后从小到大开始递归,递归的参数有:目标值(每次减去当前值)直到0,当前构造出来的数组,起始坐标等。

40. Combination Sum II


40. Combination Sum II

class Solution {
public:
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        sort(candidates.begin(),candidates.end());
        vector<vector<int>> res;
        vector<int> temp;
        backTrack(candidates,target,res,temp,0);
        return res;
    }
    void backTrack(vector<int>& candidates,int target,vector<vector<int>> &res,vector<int> &temp,int begin)
    {
        if(target<0)
            return;
        else if(target==0)
        {
            res.push_back(temp);
            return;
        }
        for(int i=begin;i<candidates.size()&&candidates[i]<=target;i++)
        {
            if(i>begin&&candidates[i]==candidates[i-1])
                continue;
            temp.push_back(candidates[i]);
            backTrack(candidates,target-candidates[i],res,temp,i+1);
            temp.pop_back();
        }
    }
};


216. Combination Sum III

class Solution {
public:
    vector<vector<int>> combinationSum3(int k, int n) {
        vector<vector<int>> res;
        vector<int> temp;
        backTrack(k,n,res,temp,1);
        return res;
    }
    void backTrack(int depth,int target,vector<vector<int>>&res,vector<int> &temp,int begin)
    {
        if(depth<0||target<0)
            return;
        else if(depth==0&&target==0)
        {
            res.push_back(temp);
            return;
        }
        for(int i=begin;i<=9&&i<=target;i++)   
        {
            temp.push_back(i);
            backTrack(depth-1,target-i,res,temp,i+1);
            temp.pop_back();
        }
    }
};

本题要递归的还有深度,因为题目要求三个数字相加,那么递归深度必须为3.