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.輸入一個不含重複數字的候選的數字集(C)和一個目標數字(T)。咱們須要找出c中的數字的不一樣組合,使得每一種組合的元素加和爲T。code
For example, 輸入的候選集[2, 3, 6, 7]和目標數字7,
結果集是:
[[7],[2, 2, 3]]遞歸
public List<List<Integer>> combinationSum(int[] nums, int target) { List<List<Integer>> res = new ArrayList<List<Integer>>(); Arrays.sort(nums); backtrack(res,new ArrayList<>(),nums,target,0); return res; } public void backtrack(List<List<Integer>> res,List<Integer> temp,int[] nums,int remain,int start){ if(remain <0)return; if(remain == 0)res.add(new ArrayList<>(temp)); else{ for(int i=start;i<nums.length;i++){ temp.add(nums[i]); backtrack(res,temp,nums,remain-nums[i],i); temp.remove(temp.size()-1); } } }