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:
target) will be positive integers.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] ]
AC code:
class Solution {
public:
vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
set<vector<int>> s;
vector<int> combination;
vector<vector<int>> res;
sort(candidates.begin(), candidates.end());
backtracking(candidates, s, combination, target, 0);
set<vector<int>>::iterator it;
for (it = s.begin(); it != s.end(); ++it)
res.push_back(*it);
return res;
}
void backtracking(vector<int>& candidates, set<vector<int>>& s, vector<int>& combination, int target, int begin) {
if (!target) {
s.insert(combination);
return ;
}
for (int i = begin; i != candidates.size() && target >= candidates[i]; ++i) {
combination.push_back(candidates[i]);
backtracking(candidates, s, combination, target-candidates[i], i+1);
combination.pop_back();
}
}
};
Runtime: 8 ms, faster than 70.36% of C++ online submissions for Combination Sum II.
原文：https://www.cnblogs.com/ruruozhenhao/p/9795031.html