Leetcode:Combination Sum II
2013-10-23 11:33
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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
A solution set is:
需要注意去重,这是个多重背包问题。可以统计每个数字出现的个数,此题就转化成了Combination
Sum
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
10,1,2,7,6,1,5and target
8,
A solution set is:
[1, 7]
[1, 2, 5]
[2, 6]
[1, 1, 6]
class Solution { public: vector<vector<int> > combinationSum2(vector<int> &num, int target) { // Note: The Solution object is instantiated only once and is reused by each test case. vector<vector<int> > v_results; if(num.empty() ) return v_results; sort(num.begin(),num.end() ); vector<int> result; helper(num,0,target,result,v_results); return v_results; } void helper(vector<int>& candidates,int pos,int target,vector<int>& result,vector<vector<int> >& v_results) { if(target==0) { v_results.push_back(result); return; } if(target<0 || pos>=candidates.size() ) return; //qu result.push_back(candidates[pos]); helper(candidates,pos+1,target-candidates[pos],result,v_results); result.pop_back(); //not qu int tmp=pos; while(tmp<candidates.size()-1 && candidates[tmp]==candidates[tmp+1]) tmp++; helper(candidates,tmp+1,target,result,v_results); } };
需要注意去重,这是个多重背包问题。可以统计每个数字出现的个数,此题就转化成了Combination
Sum
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