[Leetcode] Combination Sum

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.
• 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] 

思路：递归寻找所有可能。需要记录当前的路径以及当前的路径和。需要注意的是递归结束的条件，因为一个元素可以重复利用，所以当前元素可以继续调用当前元素，造成死循环。因此，当和超过给定值时应该返回。

class Solution {public:    int sum;    vector<int> trace;        void combination_helper(const vector<int>& candidates, int target,                            int pos, vector<vector<int>>& result) {        if (sum > target) return;        if (sum == target) {            result.push_back(trace);            return;        }        for (int i = pos; i < (int)candidates.size(); ++i) {            trace.push_back(candidates[i]);            sum += candidates[i];            combination_helper(candidates, target, i, result);            sum -= candidates[i];            trace.pop_back();        }    }        vector<vector<int> > combinationSum(vector<int> &candidates, int target) {        vector<vector<int>> result;        sort(candidates.begin(), candidates.end());        sum = 0;        combination_helper(candidates, target, 0, result);        return result;    }};

总结：复杂度为O(2^n).