LeetCode 40 Combination Sum II

Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

Each number in C may only be used once in the combination.

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 10,1,2,7,6,1,5 and target 8, 
A solution set is: 
[1, 7] 
[1, 2, 5] 
[2, 6] 
[1, 1, 6] 

public List<List<Integer>> combinationSum2(int[] candidates, int target) {Arrays.sort(candidates);List<List<Integer>> result = new LinkedList<List<Integer>>();List<Integer> tmp = new ArrayList<Integer>();dfs2(0, candidates, target, tmp, result);return result;}private void dfs2(int begin, int[] candidates, int rest, List<Integer> tmp, List<List<Integer>> result) {if (rest == 0)/**tmp是变动的,所以此处需要new一个*/result.add(new ArrayList<Integer>(tmp));/**分别从candidates[i]开始往下深搜,if candidates[i]==candidates[i-1],避免重复,跳过*/for (int i = begin; i < candidates.length && candidates[i] <= rest; i++) {if (i > begin && candidates[i] == candidates[i - 1]) continue;tmp.add(candidates[i]);dfs2(i + 1, candidates, rest - candidates[i], tmp, result);//搜索下个数字tmp.remove(tmp.size() - 1);//回溯}}