Friday, February 7, 2014

Combination Sum

LeetCode


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 (a1a2, , ak) must be in non-descending order. (iea1 ≤ 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: DFS
calculate each possible combination
/*
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]
*/
public class CombinationSum {
public ArrayList<ArrayList<Integer>> combinationSum(int[] candidates, int target) {
ArrayList<ArrayList<Integer>> result=new ArrayList<ArrayList<Integer>>();
if (candidates==null || candidates.length==0|| target<0){
return result;
}
Arrays.sort(candidates);
int start=0;
ArrayList<Integer> current=new ArrayList<Integer>();
buildResult(candidates, start, current, target, result);
return result;
}
private void buildResult(int[] candidates, int start, ArrayList<Integer> current, int target, ArrayList<ArrayList<Integer>> result){
if (target==0){
ArrayList<Integer> temp=new ArrayList<Integer>(current);
result.add(temp);
return;
}
for (int i=start; i<candidates.length; i++){
if (target-candidates[i]<0){
return;
}
current.add(candidates[i]);
buildResult(candidates, i,current, target-candidates[i], result );
current.remove(current.size()-1);
}
}
}

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