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]]

跟上一篇思想其实是一样的，只不过这个不能用同一水平线上即不能重复一个元素。上代码，递归自己写出来，看来还是懂了，嘿嘿。

public class Solution {
         public List
     
      
       > combinationSum2(int[] candidates, int target) {        List
       
        
         > res = new ArrayList
         
          
           >(); if (candidates.length == 0) return res; List
           
             list = new ArrayList
            
             (); Arrays.sort(candidates); findSum(candidates, list, 0, 0, target, res); return res; } private void findSum(int[] candidates, List
             
               list, int sum, int level, int target, List
              
               
                > res) { if (sum == target) { if (!res.contains(list)) res.add(new ArrayList
                
                 (list)); return; } else if (sum > target) return; else for (int i = level; i < candidates.length; i++) { list.add(candidates[i]); findSum(candidates, list, sum + candidates[i], i+1, target, res); list.remove(list.size() - 1); } }}