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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] ]
class Solution { public List<List<Integer>> combinationSum(int[] candidates, int target) { Arrays.sort(candidates); ArrayList<List<Integer>> ret = new ArrayList<List<Integer>>(); combinationSum(candidates, 0, target, new LinkedList<Integer>(), ret); return ret; } private void combinationSum(int[] candidates, int from, int target, LinkedList<Integer> current, ArrayList<List<Integer>> ret) { for(int i = from; i<candidates.length; ++i) { int c = candidates[i]; if(c > target) return; if(c == target) { LinkedList<Integer> sol = new LinkedList<Integer>(current); sol.add(c); ret.add(sol); return; } current.addLast(c); combinationSum(candidates, i, target -c, current, ret); current.removeLast(); } } }
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原文地址：http://www.cnblogs.com/neweracoding/p/5574660.html
