leetcode 40 Combination Sum II --- java

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]
]

这道题就是39题的变化版本，这里每一个数字只许出现一次，并且最后的结果不允许重复，第一次只是将上一题的代码修改了边界条件，但结果不是很理想。

public class combinationSum2 {
	public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        List<List<Integer>> result = new ArrayList<List<Integer>>();
        Arrays.sort(candidates);
        getResult(candidates,target,0,result,new ArrayList<Integer>());
        return result;
    }
	
	public void getResult( int[] candidates, int target,int pos, List<List<Integer>> result,List<Integer> ans){
	    for( int i = pos;i <candidates.length; i++){
	       if( target == candidates[i]){
	            ans.add(candidates[i]);
                result.add(new ArrayList<Integer>(ans));
                ans.remove(ans.size()-1);
	            return;
	       }
	       else if(target > candidates[i]){
	    	   
				ans.add(candidates[i]);
				getResult(candidates,target-candidates[i],i+1,result,ans);
				ans.remove(ans.size()-1);
		}else
		    return ;
	   }
	}
	/*
	 * 1.这道题和conbinationSum很相似，只不过不能使用重复的数字。
	 * 2.35+9
	 */
}

 之后发现，如果在getResult中，用数组代替List<Integer>那么会快很多，修改之后，做到了最快。

public class Solution {
    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        List<List<Integer>> result = new ArrayList<List<Integer>>();
        Arrays.sort(candidates);
        getResult(candidates,target,0,result,new int[candidates.length],0);
        return result;
    }
	
	public void getResult( int[] candidates, int target,int pos, List<List<Integer>> result,int[] ans,int num){
	    for( int i = pos;i <candidates.length; i++){
	       if( target == candidates[i]){
	            List<Integer> aa = new ArrayList<Integer>();
	            for( int ii =0; ii<num; ii++)
	                aa.add(ans[ii]);
	            aa.add(candidates[i]);
                result.add(aa);
	            return;
	       }
	       else if(target > candidates[i]){
				ans[num] = candidates[i];
				getResult(candidates,target-candidates[i],i+1,result,ans,num+1);
				while( i+1< candidates.length && candidates[i] == candidates[i+1])
				    i++;
			
		}else
		    return ;
	   }
}
}