leetcode_40 组合总数II

1. 题意

找到所有可能的组合数，只是不能重复选择同一元素。

组合总数II

2. 题解

2.1 我的解法

在leetcode39的基础上，再加上一个标记数组即可。

class Solution {
public:
    void gen(vector<vector<int>> &res, vector<int>& candidates, vector<int> &vis, 
        vector<int> &seq, int target)
        {
            if (target == 0) {
                res.push_back(seq);
                return ;
            }
            if ( target < 0)
                return;

            int sz = candidates.size();
            for ( int i = 0;i < sz; ++i) {
                
                if (vis[i]) continue;
                if (i && !vis[i - 1] && candidates[i] == candidates[i - 1])
                    continue;
                if ( !seq.empty() && candidates[i] < seq[seq.size() - 1] )
                    continue;
                
                vis[i] = 1;
                seq.push_back(candidates[i]);
                gen(res, candidates, vis, seq, target - candidates[i]);
                vis[i] =  0;
                seq.pop_back();


            }
        }

    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        

        vector< vector<int> > ans;
        vector<int> seq;
        vector<int> vis(candidates.size(), 0);

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

        gen(ans, candidates, vis, seq, target);
        
        return ans;
    }
};
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2.2 另一种可能的解法

其实不需要标记数组，根据有序数组的特点和begin来进行去重。

class Solution {
public:
    void gen(vector<vector<int>> &res, vector<int>& candidates, 
        vector<int> &seq, int begin, int target)
        {
            if (target == 0) {
                res.push_back(seq);
                return ;
            }
            if ( target < 0)
                return;

            int sz = candidates.size();
            for ( int i = begin;i < sz; ++i) {
                
                if ( i > begin && candidates[i] == candidates[i - 1])
                    continue;

                seq.push_back(candidates[i]);
                gen(res, candidates, seq, i + 1,target - candidates[i]);
                seq.pop_back();


            }
        }

    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        

        vector< vector<int> > ans;
        vector<int> seq;
        vector<int> vis(candidates.size(), 0);

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

        gen(ans, candidates,  seq, 0 ,target);
        
        return ans;
    }
};

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2.3 官解

统计相同数字出现的次数，一层内进行次数的枚举。

class Solution {
public:
    
    void gen(vector<vector<int>> &ans,vector<pair<int,int>> &freq,
        vector<int> &seq,int pos, int target)
    {
        if (target == 0) {
            ans.push_back(seq);
            return;
        }

        if ( pos == freq.size())
            return;

        int tm = min( target/freq[pos].first, freq[pos].second);

        for ( int i = 0; i <= tm; ++i) {
            gen(ans, freq, seq, pos + 1, target - freq[pos].first * i);
            seq.push_back(freq[pos].first);
        }

        for (int i = 0; i < tm + 1; ++i )
            seq.pop_back();

    }

    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        vector< vector<int>> ans;
        
        sort(candidates.begin(), candidates.end() );

        vector<pair<int,int>> freq;
        vector<int> seq;

        for (int v: candidates) {
            if (freq.empty() || freq.back().first != v)
                freq.push_back(make_pair(v, 1));
            else
                freq.back().second++;
        }

        gen(ans, freq, seq, 0, target);

        
        return ans;
    }
};
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