Combination Sum

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：典型的DSF题目，对每一个元素进行选择，是否需要加入。该题目中每个元素可以出现多次。因此每个元素可以多次选择。

1. public void combinationSumHelper(ArrayList<ArrayList<Integer> > res, ArrayList<Integer> cur,int[] candidates,int target,int start,int end) {
2. if(target <= 0) {
3. if(target == 0) {
4. res.add(new ArrayList<Integer>(cur)); // must add the object
5. }
6. return ;
7. }
8. for(int i=start;i<=end;i++) {
9. cur.add(candidates[i]);
10. combinationSumHelper(res,cur,candidates,target-candidates[i],i,end);
11. cur.remove(cur.size()-1);
12. }
13. }
14. public ArrayList<ArrayList<Integer>> combinationSum(int[] candidates, int target) {
15. ArrayList<ArrayList<Integer> > res = new ArrayList<ArrayList<Integer> >();
16. if(candidates==null || candidates.length==0) return res;
17. Arrays.sort(candidates);
18. ArrayList<Integer> cur = new ArrayList<Integer>();
19. combinationSumHelper(res,cur,candidates,target,0,candidates.length-1);
20. return res;
21. }

改进： 如果从当前值开始遍历，且当前值>start,且当前值 == 前一个值，则表明该值已经遍历完毕，可以直接进行下一个。

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