Given a set of candidate numbers (C) (without duplicates) 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.
The solution set must not contain duplicate combinations.
For example, given candidate set [2, 3, 6, 7] and target7,A solution set is:
[ [7], [2, 2, 3] ]
public class Solution {
private List<List<Integer>> ans = new ArrayList<List<Integer>>();
public List<List<Integer>> combinationSum(int[] candidates, int target) {
if(candidates.length==0||target<=0)
return null;
List<Integer> res = new ArrayList<Integer>();
Arrays.sort(candidates);
fun(candidates,0,target,res);
return ans;
}
private void fun(int[] candidates,int start,int target,List<Integer> curr)
{
if(target == 0)
{
List<Integer> list = new ArrayList<Integer>(curr);//若沒有這一句,直接用ans.add(curr)的話,curr每次是變化的。
ans.add(list);
}
else
{
for(int i = start;i<candidates.length;i++)
{
if(target>=candidates[i])
{
curr.add(candidates[i]);
fun(candidates,i,target-candidates[i],curr);
curr.remove(curr.size()-1);
}
}
}
}
}