问题描述:
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]
]
题源:here;完整实现:here
思路:
我们遍历所有的可能情况(使用递归的方式),然后将累加等于目标值的情况记录下来。为了防止出现重复的情况,需要先对输入数组进行排序(因为题目中输入是有序的,所以就不用了),并让我们的组合始终是升序的,代码实现如下:
vector<vector<int>> result;
void recurse(vector<int> candidates, int target, vector<int> curr){
int sum = accumulate(curr.begin(), curr.end(), 0);
if (sum == target){
result.push_back(curr); return;
}
else if (sum > target) return;
for (int i = 0; i < candidates.size(); i++){
if(curr.size() >= 1 && curr.back() > candidates[i]) continue;
else{
curr.push_back(candidates[i]);
recurse(candidates, target, curr);
curr.pop_back();
}
}
return;
}
vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
vector<int> curr;
recurse(candidates, target, curr);
return result;
}