Combination Sum
描述
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 () must be in non-descending order. (ie, ).
- 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]
分析
无
代码
// Combination Sum
// 时间复杂度O(n!),空间复杂度O(n)
class Solution {
public:
vector<vector<int> > combinationSum(vector<int> &nums, int target) {
sort(nums.begin(), nums.end());
vector<vector<int> > result; // 最终结果
vector<int> path; // 中间结果
dfs(nums, path, result, target, 0);
return result;
}
private:
void dfs(vector<int>& nums, vector<int>& path, vector<vector<int> > &result,
int gap, int start) {
if (gap == 0) { // 找到一个合法解
result.push_back(path);
return;
}
for (size_t i = start; i < nums.size(); i++) { // 扩展状态
if (gap < nums[i]) return; // 剪枝
path.push_back(nums[i]); // 执行扩展动作
dfs(nums, path, result, gap - nums[i], i);
path.pop_back(); // 撤销动作
}
}
};