Leetcode_Combination Sum 回溯法

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.