LC39 Combination Sum

Posted 2020-07-03 vaevaevae

还是利用深搜的思想，注意一个元素可以取无数次。而LC40 Combination Sum II 就有次数限制，有次数限制的情况下可以先判断一个数是否与它前面的数相等，若相等则跳过该元素，直到找到一个数与它前面的数不等，再进行深搜。

这里附上LC39的代码

 1 class Solution {
 2 public:
 3     vector<vector<int>> result;
 4     void dfs(int start,int n,int target,vector<int>& cand,vector<int> record)
 5     {
 6         int tar;
 7         for(int i=start;i<n;i++)
 8         {
 9             tar=target;
10             vector<int> rec=record;
11             if(tar<cand[i])
12                 return;
13             while(tar>=cand[i])
14             {
15                 tar-=cand[i];
16                 rec.push_back(cand[i]);
17                 if(tar==0)
18                 {
19                     result.push_back(rec);
20                     break;
21                 }
22                 else if(tar<cand[i])
23                     break;
24                 dfs(i+1,n,tar,cand,rec);
25             }
26         }
27     }
28     vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
29         if(candidates.size()==0||target<=0)
30             return result;
31         sort(candidates.begin(),candidates.end());
32         vector<int> record;
33         dfs(0,candidates.size(),target,candidates,record);
34         return result;
35     }
36 };

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附上一段更简洁的代码

 1 class Solution {
 2 public:
 3     vector<vector<int> > combinationSum(vector<int> &nums, int target) {
 4         sort(nums.begin(), nums.end());
 5         vector<vector<int> > result;  
 6         vector<int> intermediate;  
 7         dfs(nums, target, 0, intermediate, result);
 8         return result;
 9     }
10 private:
11     void dfs(vector<int>& nums, int gap, int start, vector<int>& intermediate,
12         vector<vector<int> > &result) {
13         if (gap == 0) {  
14             result.push_back(intermediate);
15             return;
16         }
17         for (size_t i = start; i < nums.size(); i++) { 
18             if (gap < nums[i]) return; 
19             intermediate.push_back(nums[i]); 
20             dfs(nums, gap - nums[i], i, intermediate, result);
21             intermediate.pop_back(); 
22         }
23     }
24 };

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