1. 题目
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.
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]
]
2. 思路
递归的思路。
每次找到一个,然后target减去之后继续找到子集再拼接起来。
如果当前的直接满足就加入到返回结果中。
为了避免重复,递归向下查找时,新的数字必须不能在当前数字的前面即可。
3. 代码
耗时:33ms
class Solution {
public:
vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
return combinationSum(candidates, target, 0);
}
vector<vector<int>> combinationSum(vector<int>& candidates, int target, int beg) {
vector<vector<int>> ret;
if (target <= 0) {return ret;}
for (int i = beg; i < candidates.size(); i++) {
int iv = candidates[i];
vector<int> one;
if (iv == target) {
one.push_back(iv);
ret.push_back(one);
}
vector<vector<int>> suf = combinationSum(candidates, target-iv, i);
for (int j = 0; j < suf.size(); j++) {
suf[j].push_back(iv);
ret.push_back(suf[j]);
}
}
return ret;
}
};
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