Combination Sum & Combination Sum II & Combination Sum III

Posted 2020-09-21 xpp

方法：采用递归的方式

class Solution {
public:
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        vector<vector<int>> result;
        sort(candidates.begin(), candidates.end());
        
        vector<int> path;
        dfs(candidates, 0, target, path, result);
        return result;
    }
    
private:
    void dfs(vector<int> &candidates, int start, int target, vector<int> &path, vector<vector<int>> &result)
    {
        if(target == 0)
            result.push_back(path);
        else
        {
            for(int i=start; i<candidates.size(); ++i)
            {
                if(candidates[i] > target) 
                    return;
                path.push_back(candidates[i]);
                dfs(candidates, i, target-candidates[i], path, result);
                path.pop_back();
            }
        }
    }
};

Combination Sum II

方法：每次递归中当前数据只访问一次，这里面需要注意一个细节问题：当原始vector中存在重复数字时，最终的结果会出现重复

class Solution {
public:
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        vector<vector<int>> result;
        vector<int> path;
        sort(candidates.begin(), candidates.end());
        
        dfs(candidates, target, path, result, 0);
        
        sort(result.begin(), result.end());
        auto iter = unique(result.begin(), result.end());
        result.erase(iter, result.end());
        return result;
    }
    
private:
    void dfs(vector<int> &candidates, int target, vector<int> &path, vector<vector<int>> &result, int start)
    {
        if(target == 0)
        {
            result.push_back(path);
            return;
        }
        else
        {
            for(int i=start; i<candidates.size(); ++i)
            {
                if(candidates[i] > target)
                    return;
                path.push_back(candidates[i]);
                dfs(candidates, target-candidates[i], path, result, i+1);
                path.pop_back();
            }
        }
    }
};

也可以使用一个简单的flag变量

class Solution {
public:
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        vector<vector<int>> result;
        vector<int> path;
        sort(candidates.begin(), candidates.end());
        
        dfs(candidates, target, path, result, 0);
        
        return result;
    }
    
private:
    void dfs(vector<int> &candidates, int target, vector<int> &path, vector<vector<int>> &result, int start)
    {
        if(target == 0)
        {
            result.push_back(path);
            return;
        }
        else
        {
            int prev = -1;
            for(int i=start; i<candidates.size(); ++i)
            {
                if(candidates[i] == prev)
                    continue;
                if(candidates[i] > target)
                    return;
                path.push_back(candidates[i]);
                prev = candidates[i];
                dfs(candidates, target-candidates[i], path, result, i+1);
                path.pop_back();
            }
        }
    }
};

 Combination Sum III

方法同上

class Solution {
public:
    vector<vector<int>> combinationSum3(int k, int n) {
        vector<vector<int>> result;
        vector<int> path;
        dfs(k, n, path, result, 1);
        return result;
    }
    
private:
    void dfs(int k, int n, vector<int> &path, vector<vector<int>> &result, int start)
    {
        if(path.size() == k && n == 0)
            result.push_back(path);
        else
        {
            for(int i=start; i<10; ++i)
            {
                if(i > n || path.size() > k)
                    return;
                path.push_back(i);
                dfs(k, n-i, path, result, i+1);
                path.pop_back();
            }
        }
    }
};