Leetcode39. Combination Sum
题目:Given a set of candidate numbers (C) (without duplicates) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.
The same repeated number may be chosen from C unlimited number of times.
Note:
- All numbers (including target) will be positive integers.
- The solution set must not contain duplicate combinations.
For example, given candidate set [2, 3, 6, 7]
and target 7
,
A solution set is:
[ [7], [2, 2, 3] ]
题意分析:这道题,选择用回溯的方法去做。采用的方法也是类似揹包问题,先假设放进去,不行再取出来。
void combinationSum(std::vector<int> &candidates, int target, std::vector<std::vector<int> > &res, std::vector<int> &combination, int begin) {
if (!target) {
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);
combination.pop_back();
}
}
代码:
class Solution {
public:
std::vector<std::vector<int> > combinationSum(std::vector<int> &candidates, int target) {
std::sort(candidates.begin(), candidates.end());
std::vector<std::vector<int> > res;
std::vector<int> combination;
combinationSum(candidates, target, res, combination, 0);
return res;
}
private:
void combinationSum(std::vector<int> &candidates, int target, std::vector<std::vector<int> > &res, std::vector<int> &combination, int begin) {
if (!target) {
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);
combination.pop_back();
}
}
};