[Leetcode 99] 40 Combination Sum II

Problem:

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.
• 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 10,1,2,7,6,1,5 and target 8, 
A solution set is: 
[1, 7] 
[1, 2, 5] 
[2, 6] 
[1, 1, 6] 

Analysis:

This is a dfs problem. First sort the given array to keep the element ordered. Then use dfs to search for candidate solutions. 

After get a solution that sum is equal to taget. Before push it into the result vector. first need to check whether there is already a duplicated solution in the result vector. If not, then we are safe to add it into the result vector.

Code:

class Solution {
public:
vector<vector<int> > res;
vector<int> t_res;
int t;

vector<vector<int> > combinationSum2(vector<int> &num, int target) {
    // Start typing your C/C++ solution below
    // DO NOT write int main() function
    res.clear();
    t_res.clear();
    t = target;

    if (num.size() == 0) return res;

    sort(num.begin(), num.end());

    dfs(0, 0, num);

    return res;
}
    
private:
    bool dup() {

    for (int i=0; i<res.size(); i++) {
        if(res[i].size() == t_res.size()) {
            int j;
            for (j=0; j<res[i].size(); j++) {
                if (res[i][j] != t_res[j])
                    break;
            }

            if (j == res[i].size())
                return true;
        }
    }

    return false;
}

void dfs(int idx, int sum, vector<int> &num) {
    if (sum == t && !dup()) {
        res.push_back(t_res);
        return ;
    } else if (sum > t)
        return ;

    for (int i=idx; i<num.size(); i++) {
        t_res.push_back(num[i]);
        dfs(i+1, sum+num[i], num);
        t_res.pop_back();
    }
}
};