Combination Sum I&&II(经典的回溯算法题)

I:

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, … , a_k) must be in non-descending order. (ie, a₁ ≤ a₂ ≤ … ≤ a_k).
• 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]

这里用到了回溯的方法，回溯其实就是一种深度优先搜索算法，相当于在整个解空间搜索问题的解，类似于穷举法，但是与穷举法的区别在于回溯法用到了剪枝，使得许多不是问题的解提前排出了，减少搜索的次数和时间。

class Solution {
private:
    vector<vector<int>> res;
    vector<int> temp;
    int tempsum=0;
   
    
public:
    void combinationSum(vector<int>& candidates, int target,vector<int>::iterator initer,int tempsum)
    {
        
      if(initer==candidates.end()||tempsum>target)
            return ;
      if(tempsum==target)
      {
           // temp.push_back(*initer);
            res.push_back(temp);
            return ;
      }
       
      for(vector<int>::iterator iter=initer;iter!=candidates.end();iter++)
      {
            temp.push_back(*iter);
            combinationSum(candidates,target,iter, tempsum+*iter);
            temp.pop_back();
           
     }
    }
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        sort(candidates.begin(),candidates.end());
        vector<int>::iterator initer=candidates.begin();
        combinationSum(candidates,target,initer,0);
        return res;
        
        
    }
};

 II:

Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

Each number in C may only be used once in the combination.

Note:

• All numbers (including target) will be positive integers.
• Elements in a combination (a₁, a₂, … , a_k) must be in non-descending order. (ie, a₁ ≤ a₂ ≤ … ≤ a_k).
• The solution set must not contain duplicate combinations.

For example, given candidate set 10,1,2,7,6,1,5 and target 8, 
A solution set is: 
[1, 7] 
[1, 2, 5] 
[2, 6] 
[1, 1, 6]

感觉自己代码写的有点复杂，不过好歹是通过了，接下来需要慢慢的把代码写简洁点，通用点。

class Solution {
private:
    vector<vector<int>> res;
    vector<int> temp;
public:
    void combinationSum(vector<int>& candidates, int target,vector<int>::iterator initer,int tempsum)
    {
      if(tempsum==target)
      {
            if(find(res.begin(),res.end(),temp)==res.end())
                res.push_back(temp);
            return ;
      }
      if(initer==candidates.end()||tempsum>target)
            return ;
      for(vector<int>::iterator iter=initer;iter!=candidates.end();iter++)
      {
            temp.push_back(*iter);
            combinationSum(candidates,target,iter+1, tempsum+*iter);
            temp.pop_back();
           
     }
    }
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        sort(candidates.begin(),candidates.end());
        vector<int>::iterator initer=candidates.begin();
        combinationSum(candidates,target,initer,0);
        return res;
        
        
    }
};