Leetcode: Combination Sum

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] 

又是DFS，熟悉了之后就比较简单了。

class Solution {public:    vector<vector<int> > combinationSum(vector<int> &candidates, int target) {        sort(candidates.begin(), candidates.end());                vector<int> v;        combinationSumUtil(candidates, v, target, 0);                return result;    }        void combinationSumUtil(vector<int> &candidates, vector<int> &comb, int sum, int start) {        if (sum == 0) {            result.push_back(comb);            return;        }                for (int i = start; i < candidates.size(); ++i) {            if (candidates[i] <= sum) {                comb.push_back(candidates[i]);                combinationSumUtil(candidates, comb, sum - candidates[i], i);                comb.pop_back();            }            else {                break;            }        }    }private:    vector<vector<int> > result;};

======================第二次======================

class Solution {public:    vector<vector<int> > combinationSum(vector<int> &candidates, int target) {        vector<vector<int>> result;        vector<int> comb;        sort(candidates.begin(), candidates.end());        combine_util(result, candidates, comb, target, 0);                return result;    }        void combine_util(vector<vector<int>> &result, vector<int> &candidates, vector<int> &comb, int target, int start) {        if (target == 0) {            result.push_back(comb);            return;        }                for (int i = start; i < candidates.size(); ++i) {            if (candidates[i] <= target) {                comb.push_back(candidates[i]);                combine_util(result, candidates, comb, target - candidates[i], i);                comb.pop_back();            }            else {                break;            }        }    }};