LeetCode: Combinations

初始系数没弄好，改了一次

class Solution {
public:
    void dfs(vector<vector<int>> &ret, int n, int k, int cur, int dep, vector<int> &tmp) {
        if (dep == k) {
            ret.push_back(tmp);
            return;
        }
        for (int i = cur; i <= n; i++) {
            tmp.push_back(i);
            dfs(ret, n, k, i+1, dep+1, tmp);
            tmp.pop_back();
        }
    }
    vector<vector<int> > combine(int n, int k) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<vector<int>> ret;
        vector<int> tmp;
        if (!k || !n || k > n) return ret;
        dfs(ret, n, k, 1, 0, tmp);
        return ret;
    }
};

 加一个iterative的方法

class Solution {
public:
    vector<vector<int> > combine(int n, int k) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        vector<vector<int> > res;
        if (k == 0) return res;
        queue<vector<int> > S;
        for (int i = 1; i <= n; i++) S.push(vector<int>(1, i));
        while (!S.empty()) {
            vector<int> tmp = S.front();
            S.pop();
            if (tmp.size() == k) res.push_back(tmp);
            else {
                for (int i = tmp.back()+1; i <= n; i++) {
                    vector<int> tmp1 = tmp;
                    tmp1.push_back(i);
                    S.push(tmp1);
                }
            }
        }
        return res;
    }
};

 C#

public class Solution {
    public List<List<int>> Combine(int n, int k) {
        List<List<int>> ans = new List<List<int>>();
        if (k == 0) return ans;
        Queue<List<int>> S = new Queue<List<int>>();
        for (int i = 1; i <= n; i++) S.Enqueue(new List<int>{i});
        while (S.Count != 0) {
            List<int> tmp = S.Peek();
            S.Dequeue();
            if (tmp.Count == k) ans.Add(tmp);
            else {
                for (int i = tmp[tmp.Count-1]+1; i <= n; i++) {
                    List<int> newTmp = new List<int>();
                    for (int j = 0; j < tmp.Count; j++) newTmp.Add(tmp[j]);
                    newTmp.Add(i);
                    S.Enqueue(newTmp);
                }
            }
        }
        return ans;
    }
}

 Java

注意clone对于基础数据类型和泛型类型来说就是deep copy，对于对象来说，对象里的基础数据类型是deep copy，而泛型类型则不是，需要override clone()。这题因为直接对List<Integer>进行clone，因此不需要override

public class Solution {
    private List<List<Integer>> ans;
    public List<List<Integer>> combine(int n, int k) {
        ans = new LinkedList<List<Integer>>();
        LinkedList<Integer> tmp = new LinkedList<Integer>();
        dfs(n, k, 1, 0, tmp);
        return ans;
    }
    private void dfs(int n, int k, int cur, int dep, LinkedList<Integer> tmp)
    {
        if (dep == k)
        {
            ans.add((List<Integer>)tmp.clone());
            return;
        }
        for (int i = cur; i <= n; i++)
        {
            tmp.add(new Integer(i));
            dfs(n, k, i+1, dep+1, tmp);
            tmp.removeLast();
        }
    }
}