leetcode 39. Combination Sum

Given a set of candidate numbers (candidates) (without duplicates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.

The same repeated number may be chosen from candidates unlimited number of times.

Note:

  • All numbers (including target) will be positive integers.
  • The solution set must not contain duplicate combinations.

Example 1:

Input: candidates = [2,3,6,7], target = 7,
A solution set is:
[
  [7],
  [2,2,3]
]

Example 2:

Input: candidates = [2,3,5], target = 8,
A solution set is:
[
  [2,2,2,2],
  [2,3,3],
  [3,5]
]

思路:回溯法

 1 class Solution {
 2     void combinationSum1(vector<int> candidates, int target, vectorint> > &res, vector<int> &v, int len, int begin) {
 3         if (target == 0) {
 4             res.push_back(v);
 5             return ;
 6         }
 7         for (int i = begin; (i < len) && target >= candidates[i]; i++) {
 8             v.push_back(candidates[i]);
 9             combinationSum1(candidates, target - candidates[i], res, v, len, i);
10             v.pop_back();
11         }
12         
13     }
14 public:
15     vectorint>> combinationSum(vector<int>& candidates, int target) {
16         sort(candidates.begin(), candidates.end());
17         vectorint> > res; 
18         vector<int> v;
19         int len = candidates.size();
20         combinationSum1(candidates, target, res, v, len, 0);
21         return res;
22     }
23 }; 

 

你可能感兴趣的:(leetcode 39. Combination Sum)