39/40 Combination Sum(I/II)

7243 ワード
LeetCode C++ 面接試験 二叉の木 回想法
Total Accepted: 83289 
Total Submissions: 274461 
Difficulty: Medium
Given a set of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.
The same repeated number may be chosen from C unlimited number of times.
Note:
All numbers (including target) will be positive integers.
Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
The solution set must not contain duplicate combinations.
For example, given candidate set  2,3,6,7  and target  7 ,  A solution set is:  [7]   [2, 2, 3]  
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Array Backtracking
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(M) Letter Combinations of a Phone Number (M) Combination Sum II (M) Combinations (M) Combination Sum III (M) Factor Combinations
分析:
本題は前の3つのブログのテーマとほぼ同じ意味ですが、ここの組み合わせは同じ数字でいいです.
本題連動:
216. Combination Sum III
77. Combinations
78/90 Subsets (I/II) 
class Solution {
public:
    void dfs(vector<int>& nums, vector<int> &subres,int sum, int start, int target)//    ，              
    {    
        if(sum == target)//      ，        
        {    
            result.push_back(subres);  
            return;  
        }
        
        for (int i = start; i < nsize; i++)    
        {    
            if(nums[i] > target  || (sum+nums[i]) > target)//       ，  
                return;
            subres.push_back(nums[i]);
            dfs(nums, subres,sum+nums[i],i,target);//          ，    。
            subres.pop_back();//                    
        }  
    }  
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        nsize=candidates.size();  
        if ( nsize == 0)     
            return result;   
        sort(candidates.begin(),candidates.end()); // ，        
        vector<int> subres;    
        dfs(candidates, subres,0,0, target);    
        return result;    
    }  
      
private:  
    vector<vector<int > > result;  
    int nsize;  
};

以前の誤った答えは、どんなに痛い悟りだろうか.
class Solution {
public:
    void dfs(vector<int>& nums, vector<int> &subres,int sum, int start, int target)//    ，              
    {    
        if(sum == target)//      ，        
        {    
            result.push_back(subres);  
            return;  
        }
        if(sum > target)//        ，  
        {   
            return;
        }
            
        for (int i = start; i < nsize; i++)    
        {    
            sum+=nums[i];//  ！！！
            subres.push_back(nums[i]);
            dfs(nums, subres,sum,i,target);
            subres.pop_back();
        }  
    }  
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        nsize=candidates.size();  
        if ( nsize == 0)     
            return result;   
        sort(candidates.begin(),candidates.end()); // ，        
        vector<int> subres;    
        dfs(candidates, subres,0,0, target);    
        return result;    
    }  
      
private:  
    vector<vector<int > > result;  
    int nsize;  
};

Total Accepted: 62935 
Total Submissions: 231780 
Difficulty: Medium
Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.
Each number in C may only be used once in the combination.
Note:
All numbers (including target) will be positive integers.
Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
The solution set must not contain duplicate combinations.
For example, given candidate set  10,1,2,7,6,1,5  and target  8 ,  A solution set is:  [1, 7]   [1, 2, 5]   [2, 6]   [1, 1, 6]  
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Hide Tags
 
Array Backtracking
Hide Similar Problems
 
(M) Combination Sum
分析:
上のテーマと区別しないで、もう分析しません!
所要時間16 ms
class Solution {
public:
    void dfs(vector<int>& nums, vector<int> &subres,int sum, int start, int target)//    ，                
    {      
        if(sum == target)//      ，          
        {      
            if(!isSameVec(subres))
                result.push_back(subres);    
            return;    
        }  
        for (int i = start; i < nsize; i++)      
        {      
            if(nums[i]>target || (sum+nums[i]) > target)//  
                return;
            subres.push_back(nums[i]);  
            dfs(nums, subres,sum+nums[i],i+1,target);//          ，    。           （       sum）  
            subres.pop_back();//                      
        }    
    }   
    bool isSameVec(vector<int> &sub)
    {
        for(int i=0;i<result.size();i++)
        {
            if(result[i]==sub)
                return true;
        }
        return false;
    }
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        nsize=candidates.size();    
        if ( nsize == 0)       
            return result;     
        sort(candidates.begin(),candidates.end()); // ，          
        vector<int> subres;      
        dfs(candidates, subres,0,0, target);      
        return result;      
    }
    
private:    
    vector<vector<int > > result;    
    int nsize; 
};

他人の優秀なアルゴリズムを学ぶ:8 ms
class Solution {
public:
    vector<vector<int> > combinationSum2(vector<int> &num, int target) 
    {
        sort(num.begin(),num.end());
        vector<int> subresult;
        dfs( 0, target, subresult, num);
        return result;
    }
    void dfs(const int order, const int target, vector<int>& subresult, const vector<int>& num)
    {
        if(target==0)
        {
            result.push_back(subresult);
            return;
        }
        
        for(int i = order;i<num.size();i++) // iterative component
        {
            if(num[i]>target) 
                return;
            if(i&&num[i]==num[i-1] && i>order) 
                continue; // check duplicate combination
            subresult.push_back(num[i]),
            dfs(i+1,target-num[i],subresult,num); // recursive componenet
            subresult.pop_back();
        }
    }
    
private:    
    vector<vector<int > > result;    
};

注:本博文はEbowTangオリジナルで、その後も本論文を更新する可能性があります.転載する場合は、必ずこの情報をコピーしてください!
原文住所:http://blog.csdn.net/ebowtang/article/details/50853069
原作者ブログ:http://blog.csdn.net/ebowtang
このブログLeetCodeの問題解索引:http://blog.csdn.net/ebowtang/article/details/50668895
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