Problem:
Given a set of candidate numbers (candidates) (without duplicates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.
The same repeated number may be chosen from candidates unlimited number of times.
Note:
- All numbers (including
target) will be positive integers. - The solution set must not contain duplicate combinations.
Example 1:
Input: candidates = [2,3,6,7], target = 7,
A solution set is:
[
[7],
[2,2,3]
]
Example 2:
Input: candidates = [2,3,5], target = 8,
A solution set is:
[
[2,2,2,2],
[2,3,3],
[3,5]
]
思路：
采用回溯法(BackTracking)。
Solution (C++):
public:
vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
sort(candidates.begin(), candidates.end());
vector<vector<int>> res;
vector<int> combination;
combinationSum(candidates, target, res, combination, 0);
return res;
}
private:
void combinationSum(vector<int> &candidates, int target, vector<vector<int>> &res, vector<int> &combination, int begin) {
if (target == 0) {
res.push_back(combination);
return ;
}
for (int i = begin; i < candidates.size() && target >= candidates[i]; ++i) {
combination.push_back(candidates[i]);
combinationSum(candidates, target - candidates[i], res, combination, i); //不是解则无变化，pop掉尾部的数
combination.pop_back(candidates[i]);
}
}
性能：
Runtime: 12 ms Memory Usage: 9.5 MB
