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] 

 

思路： 有规律的查找，避免重复。用递归得到所有的解

class Solution {public:    vector
     
      
        > combinationSum(vector
       
         &candidates, int target) {        vector
        
         
          > ans;        if(candidates.empty())            return ans;        sort(candidates.begin(), candidates.end()); //从小到大排序        recursion(ans, candidates, 0 , target);        return ans;    }    void recursion( vector
          
           
             > &ans, vector
            
              candidates, int k, int target) { static vector
             
               partans; if(target == 0) //如果partans中数字的总和已经达到目标， 压入答案 { ans.push_back(partans); return; } if(target < 0) return; for(int i = k; i < candidates.size(); i++) //当前压入大于等于candidates[k]的数字 { int sum = candidates[i]; while(sum <= target) //数字可以压入多次，只要和小于等于目标即可 { partans.push_back(candidates[i]); recursion(ans, candidates, i + 1, target - sum); //后面只压入大于当前数字的数，避免重复 sum += candidates[i]; } while(!partans.empty() && partans.back() == candidates[i]) //状态还原 partans.pop_back(); } }};
             
            
           
          
         
        
       
      
     

 

其他人更短的递归，用参数来传partans. 省略了状态还原的代码，数字压入多次也采用了递归而不是循环

class Solution {public:    void search(vector
     
      & num, int next, vector
      
       & pSol, int target, vector
       
        
          >& result)    {        if(target == 0)        {            result.push_back(pSol);            return;        }        if(next == num.size() || target - num[next] < 0)            return;        pSol.push_back(num[next]);        search(num, next, pSol, target - num[next], result);        pSol.pop_back();        search(num, next + 1, pSol, target, result);    }    vector
         
          
            > combinationSum(vector
           
             &num, int target) { vector
            
             
               > result; sort(num.begin(), num.end()); vector
              
                pSol; search(num, 0, pSol, target, result); return result; }};
              
             
            
           
          
         
        
       
      
     

 

其他人动态规划的代码，还没看，速度并不快， 但很短

class Solution {public:    vector
     
      
        > combinationSum(vector
       
         &candidates, int target) {    sort(candidates.begin(), candidates.end());    vector< vector< vector
        
          > > combinations(target + 1, vector
         
          
           >()); combinations[0].push_back(vector
           
            ()); for (auto& score : candidates) for (int j = score; j <= target; j++){ auto sls = combinations[j - score]; if (sls.size() > 0) { for (auto& s : sls) s.push_back(score); combinations[j].insert(combinations[j].end(), sls.begin(), sls.end()); } } return combinations[target];}};