19.2.3 [LeetCode 40] Combination Sum II
阿新 • • 發佈:2019-02-04
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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] ]
1 class Solution { 2 public: 3 void getsum(set<vector<int>>&res,vector<int> ans,vector<int>&candidates, int target, int lastidx) { 4 if (target == 0) { 5 res.insert(ans); 6 return; 7 } 8 if (target < candidates[lastidx])returnView Code; 9 for (int i = lastidx+1; i < candidates.size(); i++) { 10 if (target < candidates[i])break; 11 ans.push_back(candidates[i]); 12 getsum(res, ans, candidates, target - candidates[i], i); 13 ans.pop_back(); 14 } 15 } 16 vector<vector<int>> combinationSum2(vector<int>& candidates, int target) { 17 sort(candidates.begin(), candidates.end()); 18 set<vector<int>>res; 19 vector<int>ans; 20 getsum(res, ans, candidates, target, -1); 21 return vector<vector<int>>(res.begin(),res.end()); 22 } 23 };
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19.2.3 [LeetCode 40] Combination Sum II