Description
Given an array of distinct integers candidates and a target integer target, return a list of all unique combinations of candidates where the chosen numbers sum to target. You may return the combinations in any order.
The same number may be chosen from candidates an unlimited number of times. Two combinations are unique if the frequency of at least one of the chosen numbers is different.
The test cases are generated such that the number of unique combinations that sum up to target is less than 150 combinations for the given input.
Example 1:
Input: candidates = [2,3,6,7], target = 7 Output: [[2,2,3],[7]] Explanation: 2 and 3 are candidates, and 2 + 2 + 3 = 7. Note that 2 can be used multiple times. 7 is a candidate, and 7 = 7. These are the only two combinations.
Example 2:
Input: candidates = [2,3,5], target = 8 Output: [[2,2,2,2],[2,3,3],[3,5]]
Example 3:
Input: candidates = [2], target = 1 Output: []
Constraints:
1 <= candidates.length <= 302 <= candidates[i] <= 40- All elements of
candidatesare distinct. 1 <= target <= 40
Code
延續 Combinations 的概念，要注意的就是 for 迴圈的 i = start 以及呼叫 helper 時要將 start 設為 i，因為 element 是可以被重複選擇的。
class Solution {
public:
vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
vector<vector<int>> answer;
vector<int> curr;
sort(candidates.begin(), candidates.end());
helper(answer, candidates, curr, 0, 0, target);
return answer;
}
void helper(vector<vector<int>> &answer, vector<int> &candidates, vector<int> curr, int currSum, int start, int target) {
if(currSum > target) return;
if(currSum == target) {
answer.push_back(curr);
return;
}
for(int i = start; i < candidates.size(); i++) {
currSum += candidates[i];
curr.push_back(candidates[i]);
// note we pass i instead of i + 1
// since the element can be reuse
helper(answer, candidates, curr, currSum, i, target);
currSum -= candidates[i];
curr.pop_back();
}
}
};