LeetCode 39. Combination Sum

Description:

Given a set of candidate numbers (C) (without duplicates) 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.

Example:

For example, given candidate set [2, 3, 6, 7] and target 7,
A solution set is:
[
[7],
[2, 2, 3]
]

代码如下：

#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
class Solution {
public:
    vector<vector<int> > combinationSum(vector<int>& candidates, int target) {
        vector<vector<int> > res;
        vector<int> combination;
        sort(candidates.begin(), candidates.end());
        combinationSum2(candidates, target, res, combination, 0);
        return res;
    }
private:
    void combinationSum2(vector<int>& candidates, int target, vector<vector<int> >& res, vector<int>& combination, int begin) {
        if (!target) {
            res.push_back(combination);
            return;
        }
        for (int i = begin; i != candidates.size() && target >= candidates[i]; i++) {
            if (i == begin || candidates[i] != candidates[i - 1]) {
                combination.push_back(candidates[i]);
                combinationSum2(candidates, target - candidates[i], res, combination, i);
                combination.pop_back();
            }
        }
    }
};
int main() {
    int n;
    vector<int> nums;
    cin >> n;
    for (int i = 0; i < n; i++) {
        int t;
        cin >> t;
        nums.push_back(t);
    }
    int target;
    cin >> target;
    Solution s;
    vector<vector<int> > res = s.combinationSum(nums, target);
    for (int i = 0; i < res.size(); i++) {
        for (int j = 0; j < res[i].size(); j++) {
            cout << res[i][j] << " ";
        }
        cout << endl;
    }
    return 0;
}

Description:

Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

Each number in C may only be used once in the combination.

Note:
All numbers (including target) will be positive integers.
The solution set must not contain duplicate combinations.

Example:

For example, given candidate set [10, 1, 2, 7, 6, 1, 5] and target 8,
A solution set is:
[
[1, 7],
[1, 2, 5],
[2, 6],
[1, 1, 6]
]

代码如下：

`C++
class Solution {
public:
vector<vector > combinationSum2(vector& candidates, int target) {
vector<vector > res;
vector combination;
sort(candidates.begin(), candidates.end());
combinationSum2(candidates, target, res, combination, 0);
return res;
}
private:
void combinationSum2(vector& candidates, int target, vector<vector >& res, vector& combination, int begin) {
if (!target) {
res.push_back(combination);
return;
}
for (int i = begin; i != candidates.size() && target >= candidates[i]; i++) {
if (i == begin || candidates[i] != candidates[i - 1]) {
combination.push_back(candidates[i]);
combinationSum2(candidates, target - candidates[i], res, combination, i + 1);
combination.pop_back();
}
}
}
};