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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]
思路：使用回溯和递归
List<List<Integer>> lists = new ArrayList<List<Integer>>(); public List<List<Integer>> combinationSum(int[] candidates, int target) { Arrays.sort(candidates); if(target == 0 || candidates == null || candidates.length == 0) { return lists; } combinationSumHelper(candidates, target, 0, new ArrayList<Integer>()); return lists; } public void combinationSumHelper(int[] candidates, int target, int start, List<Integer> list) { for(int i=start; i<candidates.length; i++) { if(target > candidates[i]) { list.add(candidates[i]); combinationSumHelper(candidates, target-candidates[i], i, list); list.remove(list.size()-1); } else if(target == candidates[i]) { list.add(candidates[i]); lists.add(new ArrayList(list)); list.remove(list.size()-1); return; } else { return; } } }
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原文地址：http://www.cnblogs.com/linxiong/p/4297402.html
