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. 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]
]

class Solution {
public:
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        //sort(candidates.begin(), candidates.end());
        vector<int> sol;
        vector<vector<int>> res;

        dfs(candidates, target, 0, sol, res);
        return res;

    }
private:
    void dfs(vector<int> &arr, int target, int level,vector<int> &sum, vector<vector<int>>& ret){
        if(target == 0){
            ret.push_back(sum);
            return;
        }

        for(int i = level; i< arr.size(); i++){
            if(arr[i] <= target){
                sum.push_back(arr[i]);
                dfs(arr, target - arr[i], i, sum, ret);
                sum.pop_back();
            }
        }
    }
};

40.Combination Sum II

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. 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]
]

注意这道题目与上面的区别是给出的数据集合可能有重复的值

class Solution {
public:
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        sort(candidates.begin(), candidates.end());
        vector<vector<int>> res;
        vector<int> sum;

        dfs(candidates, target, 0, sum, res);
        sort(res.begin(), res.end());
        auto it = unique(res.begin(), res.end());
        res.erase(it, res.end());
        return res;
    }
private:
    void dfs(vector<int> &arr, int left, int level, vector<int>& sol, vector<vector<int>> &ret){
        if(left == 0){
            sort(sol.begin(), sol.end());
            ret.push_back(sol);
            return;
        }

        for(int i = level; i < arr.size(); i++){
            if(left >= arr[i]){
                sol.push_back(arr[i]);
                dfs(arr, left - arr[i], i + 1, sol, ret);
                sol.pop_back();
            }
        }
    }
};