首页 > 代码库 > 从n个元素中选择k个的所有组合(包含重复元素)

从n个元素中选择k个的所有组合(包含重复元素)

LeetCode:Combinations这篇博客中给出了不包含重复元素求组合的5种解法。我们在这些解法的基础上修改以支持包含重复元素的情况。对于这种情况,首先肯定要对数组排序,以下不再强调

修改算法1:按照求包含重复元素集合子集的方法LeetCode:Subsets II算法1的解释,我们知道:若当前处理的元素如果在前面出现过m次,那么只有当前组合中包含m个该元素时,才把当前元素加入组合

?
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
class Solution {
public:
    void combine(vector<int> &vec, int k) {
        if(k > vec.size())return;
        sort(vec.begin(), vec.end());
 
        vector<int>tmpres;
        helper(vec, 0, k, 0, tmpres);
    }
     
    //从vec的[start,vec.size()-1]范围内选取k个数,tmpres是当前组合
    //times是上一个元素出现的次数
    void helper(vector<int> &vec, int start, int k, int times, vector<int> &tmpres)
    {
        if(vec.size()-start < k)return;
        if(k == 0)
        {
            for(int i = 0; i < tmpres.size(); i++)
                cout<<tmpres[i]<<" ";
            cout<<endl;
            return;
        }
        if(start == 0 || vec[start] != vec[start-1])//当前元素前面没有出现过
        {
            //选择vec[start]
            tmpres.push_back(vec[start]);
            helper(vec, start+1, k-1, 1, tmpres);
            tmpres.pop_back();
            //不选择vec[start]
            helper(vec, start+1, k, 1, tmpres);
        }
        else//当前元素前面出现过
        {
            if(tmpres.size() >= times && tmpres[tmpres.size()-times] == vec[start])
            {
                //只有当tmpres中包含times个vec[start]时,才选择vec[start]
                tmpres.push_back(vec[start]);
                helper(vec, start+1, k-1, times+1, tmpres);
                tmpres.pop_back();
            }
            helper(vec, start+1, k, times+1, tmpres);
        }
    }
};

从[1,2,2,3,3,4,5]中选3个的结果如下:

image


修改算法2:同理,可以得到代码如下                    本文地址

?
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
class Solution {
public:
    void combine(vector<int> &vec, int k) {
        if(k > vec.size())return;
        sort(vec.begin(), vec.end());
 
        vector<int>tmpres;
        helper(vec, 0, k, 0, tmpres);
    }
     
    //从vec的[start,vec.size()-1]范围内选取k个数,tmpres是当前组合
    //times是上一个元素出现的次数
    void helper(vector<int> &vec, int start, int k, int times, vector<int> &tmpres)
    {
        if(vec.size()-start < k)return;
        if(k == 0)
        {
            for(int i = 0; i < tmpres.size(); i++)
                cout<<tmpres[i]<<" ";
            cout<<endl;
            return;
        }
        for(int i = start; i <= vec.size()-k; i++)
        {
            if(i == 0 || vec[i] != vec[i-1])//当前元素前面没有出现过
            {
                times = 1;
                //选择vec[i]
                tmpres.push_back(vec[i]);
                helper(vec, i+1, k-1, 1, tmpres);
                tmpres.pop_back();
            }
            else//当前元素前面出现过
            {
                times++;
                //vec[i]前面已经出现过times-1次
                if(tmpres.size() >= times-1 && tmpres[tmpres.size()-times+1] == vec[i])
                {
                    //只有当tmpres中包含times-1个vec[i]时,才选择vec[i]
                    tmpres.push_back(vec[i]);
                    helper(vec, i+1, k-1, times, tmpres);
                    tmpres.pop_back();
                }
            }
        }
    }
};

修改算法3:算法3是根据LeetCode:Subsets 算法2修改未来,同理我们也修改LeetCode:SubsetsII 算法2

?
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
class Solution {
public:
    void combine(vector<int> &vec, int k) {
        if(k > vec.size())return;
        sort(vec.begin(), vec.end());
 
        vector<vector<int> > res(1);//开始加入一个空集
        int last = vec[0], opResNum = 1;//上一个数字、即将要进行操作的子集数量
        for(int i = 0; i < vec.size(); ++i)
        {
            if(vec[i] != last)
            {
                last = vec[i];
                opResNum = res.size();
            }
            //如果有重复数字,即将操作的子集的数目和上次相同
            int resSize = res.size();
            for(int j = resSize-1; j >= resSize - opResNum; j--)
            {
                res.push_back(res[j]);
                res.back().push_back(vec[i]);
                if(res.back().size() == k)//找到一个大小为k的组合
                {
                    for(int i = 0; i < res.back().size(); i++)
                        cout<<res.back()[i]<<" ";
                    cout<<endl;
                }
            }
        }
    }
};

对于算法4和算法5,都是基于二进制思想,这种解法不适用与包含重复元素的情况

 

【版权声明】转载请注明出处:http://www.cnblogs.com/TenosDoIt/p/3695463.html