[*leetcode 39] Combination Sum && [*leetcode 40] Combination Sum II

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] 

[Solution]

使用递归

 1 vector<vector<int> > result;
 2     vector<int> line;
 3     vector<vector<int> > combinationSum(vector<int> &candidates, int target) 
 4     {
 5         sort(candidates.begin(), candidates.end());
 6         combination(candidates, 0, 0, target);
 7         return result;
 8     }
 9     
10     void combination(vector<int> &candidates, int index, int sum, int target)
11     {
12         if (sum >= target)
13         {
14             if (sum == target)
15                 result.push_back(line);
16             return;
17         }
18         
19         for (int i = index; i < candidates.size(); i++)
20         {
21             line.push_back(candidates[i]);
22             combination(candidates, i, sum + candidates[i], target);
23             line.pop_back();
24         }
25     }

 

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] 

[Solution]

和上一题相比，需要每个candidate只取一次，且去重复

 1 vector<vector<int> > result;
 2     vector<int> line;
 3     vector<vector<int> > combinationSum2(vector<int> &candidates, int target) 
 4     {
 5         sort(candidates.begin(), candidates.end());
 6         combination(candidates, 0, 0, target);
 7         return result;
 8     }
 9     
10     void combination(vector<int> &candidates, int index, int sum, int target)
11     {
12         if (sum >= target)
13         {
14             if (sum == target)
15                 result.push_back(line);
16             return;
17         }
18         
19         for (int i = index; i < candidates.size(); i++)
20         {
21             if (i > index && candidates[i] == candidates[i - 1]) // delete duplicate
22                 continue;
23             line.push_back(candidates[i]);
24             combination(candidates, i + 1, sum + candidates[i], target); // only use once.
25             line.pop_back();
26         }
27     }