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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]
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++; } } }
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原文地址：http://www.cnblogs.com/aboutblank/p/4538445.html
