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:
For example, given candidate set 2,3,6,7
and target 7
,
A solution set is:
[7]
[2, 2, 3]
Solution: use the DP and recursion method.
Note the back tracking and the null pointer error in line 21.(still don't know why)
public class Solution { public ArrayList<ArrayList<Integer>> combinationSum(int[] candidates, int target) { // Note: The Solution object is instantiated only once and is reused by each test case. ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>(); int len = candidates.length; if(candidates == null || len == 0){ return result; } Arrays.sort(candidates); if(target < candidates[0]){ return result; } ArrayList<Integer> al = new ArrayList<Integer>(); findSum(candidates, 0, len, target, result, al); return result; } public void findSum(int[] candidates, int start, int end, int target, ArrayList<ArrayList<Integer>> result, ArrayList<Integer> al){ if(target == 0){ result.add(new ArrayList<Integer>(al)); //at first I wrote result.add(al) here, and get the wrong answer, don't know why. return; } for(int i=start; i<end; i++){ if(target < candidates[i]){ break; } al.add(candidates[i]); target -= candidates[i]; findSum(candidates, i, end, target, result, al); target += al.remove(al.size()-1); //use the back traking here! } } }