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] 

z这一道跟前一道意思差不多，区别在于元素不允许重复，这个只要在前面那一道的判断中，

if(sum(stack)+candidates<target)i--;//而不是i不变就可以了。

另外题目要求不能出现相同的答案，所有在最后的结果集中，

用if(!result.contains(list))result.add(list)来挑出相同的。

有了前面那道的基础，做这道题就比较简单了。

上代码

public class Solution {    public List<List<Integer>> combinationSum2(int[] candidates, int target) {        Stack<Integer> stack=new Stack<>();List<Stack<Integer>> listtmp=new ArrayList<>();//List<List<Map<Integer, Integer>>> resultList=new ArrayList<>();int sum=0;Arrays.sort(candidates);for(int i=candidates.length-1;i>=0||!stack.isEmpty();){if(i<0){if(stack.isEmpty())break;i=stack.pop();sum-=candidates[i];i--;continue;}if(candidates[i]+sum==target){stack.push(i);//Stack tobeadd=(Stack<Integer>) stack1.clone ();listtmp.add((Stack<Integer>) stack.clone ());stack.pop();i--;}elseif(candidates[i]+sum>target){i--;}elseif(candidates[i]+sum<target){sum+=candidates[i];stack.push(i);i--;}}List<List<Integer>> result=new ArrayList<>();for (Stack<Integer> keyarray : listtmp) {List<Integer> childIntegers=new ArrayList<>();for(Integer key:keyarray){childIntegers.add(0,candidates[key]);}if(!result.contains(childIntegers))result.add(childIntegers);}return result;    }}