1. Question

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

Each number in candidates 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.

Example 1:

Input: candidates = [10,1,2,7,6,1,5], target = 8, A solution set is: [ [1, 7], [1, 2, 5], [2, 6], [1, 1, 6] ]

Example 2:

Input: candidates = [2,5,2,1,2], target = 5, A solution set is: [ [1,2,2], [5] ]

2. Solution

里面注意去掉重复元素的方式： if (i > index && candidates[i] == candidates[i - 1]) continue;

class Solution {
 public:
  vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
    vector<vector<int>> res;
    vector<int> temp;
    sort(candidates.begin(), candidates.end());
    findcombinationSum(candidates, target, 0, temp, res);
    return res;
  }

  void findcombinationSum(vector<int>& candidates, int target, int index,
                          vector<int>& temp, vector<vector<int>>& res) {
    if (target < 0) return;
    if (target == 0) res.emplace_back(temp);
    for (int i = index; i < candidates.size(); ++i) {
      if (i > index && candidates[i] == candidates[i - 1]) continue;
      temp.emplace_back(candidates[i]);
      findcombinationSum(candidates, target - candidates[i], i + 1, temp, res);
      temp.pop_back();
    }
  }
};

3. Updated

• 03/01/2019 Review(BFS, 6mins)