40. Combination Sum II

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]
]

比起上个题多了个条件,结果不能重复,并且给定的元素有重复;
解决办法,在dfs的每层循环添加元素时,判断和前一个是不是相等,相等则不添加;

class Solution {
public:
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        sort(candidates.begin(),candidates.end());
        vector<vector<int>> res;
        vector<int> tmp;
        dfs(res,tmp,target,candidates,0);
        return res;
    }
    
    void dfs(vector<vector<int>> &res, vector<int> &tmp, int target, vector<int> &candidates, int index)
    {
        if(0==target)
        {
            res.push_back(tmp);
            return;
        }
        
        for(int i=index;i<candidates.size()&&candidates[i]<=target;++i) //注意加上 candidates<=targe, 否则容易超时
        {
            if(i!=index&&candidates[i]==candidates[i-1])continue;
            tmp.push_back(candidates[i]);
            dfs(res,tmp,target-candidates[i], candidates,i+1);
            tmp.pop_back();
        }
    }
};