CODE 93: 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 ArrayList<ArrayList<Integer>> combinationSum2(int[] candidates,int target) {// IMPORTANT: Please reset any member data you declared, as// the same Solution instance will be reused for each test case.Arrays.sort(candidates);return dfs(candidates, target, 0, 0);}ArrayList<ArrayList<Integer>> dfs(int[] candidates, int target, int sum,int i) {if (i >= candidates.length) {ArrayList<ArrayList<Integer>> results = new ArrayList<ArrayList<Integer>>();return results;}ArrayList<ArrayList<Integer>> results = new ArrayList<ArrayList<Integer>>();for (int index = 0; index * candidates[i] <= target && index <= 1; index++) {int tmpsum = sum + index * candidates[i];if (tmpsum == target) {ArrayList<Integer> result = new ArrayList<Integer>();for (int m = 0; m < index; m++) {result.add(0, candidates[i]);}if (!results.contains(result)) {results.add(result);}} else if (tmpsum < target) {ArrayList<ArrayList<Integer>> tmpResults = dfs(candidates,target, tmpsum, i + 1);for (ArrayList<Integer> tmpResult : tmpResults) {ArrayList<Integer> newResult = new ArrayList<Integer>();newResult.addAll(tmpResult);for (int m = 0; m < index; m++) {newResult.add(0, candidates[i]);}if (!results.contains(newResult)) {results.add(newResult);}}}}return results;}