Combination Sum II leetcode java

題目：

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]

 

題解：

這道題跟combination sum本質的差別就是當前已經遍歷過的元素只能出現一次。

所以需要給每個candidate一個visited域，來標識是否已經visited了。

當回退的時候，記得要把visited一起也回退了。

代碼如下：

 
           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);   
          
 
          10          
          return res;   
          
 
          11     }   
          
 
          12      
          
 
          13      
          private  
          static  
          void helper( 
          int[] candidates,  
          int target,  
          int start, ArrayList<Integer> item,    
          
 
          14     ArrayList<ArrayList<Integer>> res,  
          boolean[] visited){   
          
 
          15          
          if(target<0)   
          
 
          16              
          return;   
          
 
          17          
          if(target==0){   
          
 
          18             res.add( 
          new ArrayList<Integer>(item));   
          
 
          19              
          return;   
          
 
          20         } 
          
 
          21          
          
 
          22          
          for( 
          int i=start;i<candidates.length;i++){ 
          
 
          23              
          if(!visited[i]){ 
          
 
          24                  
          if(i>0 && candidates[i] == candidates[i-1] && visited[i-1]== 
          false) 
          // 
          deal with dupicate 
          
 
          25  
                               
          continue;   
          
 
          26                 item.add(candidates[i]); 
          
 
          27                 visited[i]= 
          true; 
          
 
          28                  
          int newtarget = target - candidates[i]; 
          
 
          29                 helper(candidates,newtarget,i+1,item,res,visited);   
          
 
          30                 visited[i]= 
          false; 
          
 
          31                 item.remove(item.size()-1);   
          
 
          32             } 
          
 
          33         }   
          
 
          34     }