【LeetCode】Combination Sum

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

思路：深度优先遍历，子树的根节点必须是大于等于根节点的数字，构造的树，记录下路径，直到sum大于等于target，等于就放到vector中，整个过程要借助一个path的路径变量来记录经过的节点数字。

class Solution {public:void comb(vector<int> candidates, int index, int sum, int target, vector<vector<int>> &res, vector<int> &path)      {          if(sum>target)return;          if(sum==target){res.push_back(path);return;}          for(int i= index; i<candidates.size();i++)          {              path.push_back(candidates[i]);              comb(candidates,i,sum+candidates[i],target,res,path);              path.pop_back();          }      }      vector<vector<int> > combinationSum(vector<int> &candidates, int target) {          // Note: The Solution object is instantiated only once.          sort(candidates.begin(),candidates.end());          vector<vector<int>> res;          vector<int> path;          comb(candidates,0,0,target,res,path);          return res;      }  };