【LeetCode】77. Combinations (2 solutions)

Combinations

Given two integers n and k, return all possible combinations of k numbers out of 1 ... n.

For example,
If n = 4 and k = 2, a solution is:

[

  [2,4],

  [3,4],

  [2,3],

  [1,2],

  [1,3],

  [1,4],

]

解法一：递归

递推点：加入i后，下一个加入的元素需要遍历i+1~n

因此可以基于k做递归。

base case: k==cur.size()，此时cur即为符合条件的一个集合。

class Solution {

public:

    vector<vector<int> > combine(int n, int k) {

        vector<vector<int> > ret;

        vector<int> cur;

        Helper(ret, cur, k, , n);

        return ret;

    }

    void Helper(vector<vector<int> >& ret, vector<int> cur, int k, int pos, int n)

    {

        if(cur.size() == k)

            ret.push_back(cur);

        else

        {

            for(int i = pos; i <= n; i ++)

            {

                cur.push_back(i);

                Helper(ret, cur, k, i+, n);

                cur.pop_back();

            }

        }

    }

};

解法二：非递归

遍历子集过程中，大小为k的子集即为所需集合。

注意略去大小超过k的子集，若不然存储所有子集需要2^n空间。

子集遍历法参考Subsets

class Solution {

public:

    vector<vector<int> > combine(int n, int k) {

        vector<vector<int> > ret;

        vector<vector<int> > subsets;

        vector<int> cur;    //empty set

        //check all subsets with k elements

        subsets.push_back(cur);

        for(int i = ; i <= n; i ++)

        {//all element put into all subsets in ret

            int size = subsets.size();

            for(int j = ; j < size; j ++)

            {

                cur = subsets[j];

                if(cur.size() >= k)

                    continue;

                cur.push_back(i);

                if(cur.size() == k)

                    ret.push_back(cur);

                else

                //cur.size() < k

                    subsets.push_back(cur);

            }

        }

        return ret;

    }

};

巴特西

【LeetCode】77. Combinations (2 solutions)

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