Given a set of candidate numbers (C) 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.
- 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 2,3,6,7 and target 7,
A solution set is: [7] [2, 2, 3]
解题思路：
使用回溯的思想穷举可能的结果。
代码：
class Solution {
public:
vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
vector<vector<int>> lst;
vector<int> ans;
sort(candidates.begin(), candidates.end());
Backtracking(lst, ans, candidates, , target);
return lst;
}
void Backtracking(vector<vector<int>> &lst, vector<int> ans, vector<int> candidates, int idx, int left) {
for (int i = idx; i < candidates.size(); ++i) {
int cur_left = left - candidates[i];
if (cur_left == ) {
ans.push_back(candidates[i]);
lst.push_back(ans);
return;
}
if (cur_left > ) {
ans.push_back(candidates[i]);
Backtracking(lst, ans, candidates, i, cur_left);
ans.pop_back();
} else {
return;
}
}
}
};另：是否还有DP/DFS等其他思路。