Combination Sum （without Duplication）

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.
• 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]]

代码如下：

import java.util.ArrayList;import java.util.Arrays;import java.util.List;/** * 不用相同的元素 */public class CombinationSumII {public List<ArrayList<Integer>> combinationSumII(int[] nums,int target){List<ArrayList<Integer>> list = new ArrayList<>();Arrays.sort(nums);backtrack(list,new ArrayList<>(),nums,target,0);return list;}private void backtrack(List<ArrayList<Integer>> list, ArrayList<Integer> arrayList, int[] nums, int target, int start) {if(target<0)return;else if(target == 0)list.add(arrayList);else{for(int i=start;i<nums.length;i++){if(i>start && nums[i]==nums[i-1])//跳过重复的数continue;arrayList.add(nums[i]);backtrack(list,arrayList,nums,target-nums[i],i+1);arrayList.remove(arrayList.size()-1);}}}}