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]
题目大意:给一个候选数组,一个目标值,从候选数组找出所有和等于目标值的List,候选数组元素可以重复。
解题思路:还是用回溯的方式来做,因为可以重复,所以每次都可以从当前元素开始往后依次添加到List,递归判断,然后回溯。
public List<List<Integer>> combinationSum(int[] candidates, int target) { List<List<Integer>> res = new ArrayList<>(); if (candidates == null || candidates.length == 0) { return res; } List<Integer> tmp = new ArrayList<>(); helper(res, tmp, 0,candidates, target); return res; } private void helper(List<List<Integer>> res, List<Integer> tmp, int start,int[] candidates, int target) { if (0 == target) { res.add(new ArrayList<>(tmp)); // System.out.println(tmp); return; } for (int i = start; i < candidates.length && target >= candidates[i]; i++) { tmp.add(candidates[i]); helper(res, tmp, i,candidates, target - candidates[i]); tmp.remove(tmp.size() - 1); int ca = candidates[i]; while(i<candidates.length&&ca==candidates[i]){ i++; } } }
时间: 2024-10-17 11:41:19