[leetcode-40-Combination Sum 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.
• 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]
]

思路：

典型回溯法，需要注意的就是存在重复情况的处理。

 

 

void combine(vector<vector<int>>& res, vector<int>& candidates, vector<int>& cantemp,int begin, int target)
     {
         if (target < 0) return;
         
         if ( target == 0)
         {
             res.push_back(cantemp);
             return;
         }
         for (int i = begin; i < candidates.size();i++)
         {
             //if (candidates[i]>target)return;
             if (i && candidates[i]==candidates[i-1]&& i>begin) continue;//避免重复元素
             cantemp.push_back(candidates[i]);
             combine(res, candidates, cantemp, i + 1, target - candidates[i]);
             cantemp.pop_back();             
         }
     }
     vector<vector<int>> combinationSum2(vector<int>& candidates, int target)
     {        
         vector<vector<int>> res;
         if (candidates.size() == 0)return res;
         sort(candidates.begin(), candidates.end());
         vector<int> cantemp;
         combine(res, candidates, cantemp, 0, target);
         return res;
     }

参考：

https://discuss.leetcode.com/topic/8916/c-backtracking-solution-with-detailed-explanation