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
2,3,6,7and target7,
A solution set is:
[7]
[2, 2, 3]
先对数组排序，使用递归选取元素，选择递归的条件。private void getCom(int[] candidates, int target, int start, List<Integer> subList, List<List<Integer>> list) { for (int i = start; i < candidates.length; i++) { List<Integer> newList = new ArrayList<Integer>(subList); if (target - candidates[i] == 0) { newList.add(candidates[i]); list.add(newList); } else if (target - candidates[i] > 0) { newList.add(candidates[i]); getCom(candidates, target - candidates[i], i, newList, list); } else { break; } } } public List<List<Integer>> combinationSum(int[] candidates, int target) { Arrays.sort(candidates); List<List<Integer>> list = new ArrayList<List<Integer>>(); List<Integer> subList = new ArrayList<Integer>(); getCom(candidates, target, 0, subList, list); return list; }
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原文：http://blog.csdn.net/u010378705/article/details/30301267