LeetCode 39. Combination Sum

Given a set of candidate numbers (C) and a target number (T), find all unique combinations inC 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, … ,a_k) must be in non-descending order. (ie, a₁ ≤a₂ ≤ … ≤ a_k).
• 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]




 Need to pay special attention to the condition that "The number can be used repeatedly", we thus use a pointer to pointer to the value used and not adding one when being used in the next level.

#include <vector>#include <iostream>#include <algorithm>using namespace std;void combinationSum(vector<int>& candidates, int start, vector< vector<int> >& res, vector<int>& path, int& sum, int target) {    if(sum > target) return;    if(sum == target) {        res.push_back(path);    }    for(int i = start; i < candidates.size(); ++i) {        sum += candidates[i];        path.push_back(candidates[i]);        combinationSum(candidates, i, res, path, sum, target);        path.pop_back();        sum -= candidates[i];    }}vector< vector<int> > combinationSum(vector<int>& candidates, int target) {    vector< vector<int> > res;    vector<int> path;    sort(candidates.begin(), candidates.end());    int sum = 0;    combinationSum(candidates, 0,  res, path, sum, target);    return res;}int main(void) {    vector<int> candidates{2, 3, 6, 7};    vector< vector<int> > res = combinationSum(candidates, 7);    for(int i = 0; i < res.size(); ++i) {        for(int j = 0; j < res[0].size(); ++j) {            cout << res[i][j] << endl;        }        cout << endl;    }}