标签：== nta temp ++ repeat ati bin date can
Given a set of candidate numbers (candidates) (without duplicates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.
The same repeated number may be chosen from candidates unlimited number of times.
Note:
target) will be positive integers.Example 1:
Input: candidates =[2,3,6,7],target =7, A solution set is: [ [7], [2,2,3] ]
Example 2:
Input: candidates = [2,3,5],target = 8, A solution set is: [ [2,2,2,2], [2,3,3], [3,5] ]
1 class Solution { 2 3 public void dfs(List<List<Integer>> ans, int []candidates, int pos, int target, ArrayList<Integer> temp) { 4 if (target == 0) { 5 6 7 ans.add(new ArrayList<>(temp)); 8 9 return; 10 } 11 if (target < candidates[pos]) { 12 return; 13 } 14 int n = candidates.length; 15 System.out.println(pos); 16 for (int i = pos; i < n && candidates[i] <= target; ++i) { 17 temp.add(candidates[i]); 18 dfs(ans, candidates, i, target - candidates[i], temp); 19 temp.remove(temp.size() - 1); 20 } 21 } 22 public List<List<Integer>> combinationSum(int[] candidates, int target) { 23 Arrays.sort(candidates); 24 List<List<Integer>> ans = new ArrayList<>(); 25 if (candidates.length == 0) return ans; 26 int n = candidates.length; 27 ArrayList<Integer> temp = new ArrayList<>(); 28 dfs(ans, candidates, 0, target, temp); 29 return ans; 30 } 31 }
标签：== nta temp ++ repeat ati bin date can
原文地址：https://www.cnblogs.com/hyxsolitude/p/12315463.html
