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, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
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]
vector<vector<int> > combinationSum2(vector<int> &num, int target) { vector<vector<int> >res; vector<int>temp; vector<bool>vis(num.size(),false); sort(num.begin(),num.end()); dfs(num,target,0,temp,res,vis); return res;}void dfs(vector<int>& num, int gap, int idx, vector<int>& temp, vector<vector<int> >& res, vector<bool>& vis){ if(gap == 0){ res.push_back(temp); return; } for(int i=idx; i<num.size(); i++){ if(i > 0 && num[i] == num[i-1] && !vis[i-1]) continue; if(num[i] > gap) return; temp.push_back(num[i]); vis[i] = true; dfs(num,gap-num[i],i+1,temp,res,vis); temp.pop_back(); vis[i] = false; }}
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