Combination Sum II leetcode java

题目：

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, … , a_k) must be in non-descending order. (ie, a₁ ≤ a₂ ≤ … ≤ a_k).
• 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]

题解：

这道题跟combination sum本质的差别就是当前已经遍历过的元素只能出现一次。

所以需要给每个candidate一个visited域，来标识是否已经visited了。

当回退的时候，记得要把visited一起也回退了。

代码如下：

 1     public static ArrayList<ArrayList<Integer>> combinationSum(int[] candidates, int target) {  
 2         ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();  
 3         ArrayList<Integer> item = new ArrayList<Integer>();
 4         if(candidates == null || candidates.length==0)  
 5             return res; 
 6             
 7         Arrays.sort(candidates);  
 8         boolean [] visited = new boolean[candidates.length];
 9         helper(candidates,target, 0, item ,res, visited);  
         return res;  
     }  
     
     private static void helper(int[] candidates, int target, int start, ArrayList<Integer> item,   
     ArrayList<ArrayList<Integer>> res, boolean[] visited){  
         if(target<0)  
             return;  
         if(target==0){  
             res.add(new ArrayList<Integer>(item));  
             return;  
         }
         
         for(int i=start;i<candidates.length;i++){
             if(!visited[i]){
                 if(i>0 && candidates[i] == candidates[i-1] && visited[i-1]==false)//deal with dupicate
                     continue;  
                 item.add(candidates[i]);
                 visited[i]=true;
                 int newtarget = target - candidates[i];
                 helper(candidates,newtarget,i+1,item,res,visited);  
                 visited[i]=false;
                 item.remove(item.size()-1);  
             }
         }  
     }